The heat required to raise the ice to 0 ° C .

Question

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Answer 1

Question

Chapter 10, Problem 3SP

(a)

To determine

The heat required to raise the ice to 0°C.

(a)

Expert Solution

Answer to Problem 3SP

The heat required to raise the ice to 0°C is 1540 cal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

Here,

Q is the heat

m is the mass

c is the specific heat capacity of ice

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Here,

ΔT is the change in temperature

T1 is the initial temperature

T2 is the final temperature

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 140 g for m, 0.5 cal/g⋅°C for c, −22°C for T1 and 0°C for T2 in equation (3)

Q=140 g×0.5 cal/g⋅°C(0°C−(−22°C)) =1540 cal

Conclusion:

The heat required to raise the ice to 0°C is 1540 cal.

(b)

To determine

The additional heat required to completely melt ice.

(b)

Expert Solution

Answer to Problem 3SP

The additional heat required to completely melt ice is 11.2 kcal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of latent heat capacity

Q=mLf (1)

Here,

Q is the heat required tot completely melt ice

m is the mass

Lf is the latent heat capacity

Substitute 140 g for m and 80 cal/g for Lf

Q=140 g×80 cal/g=11200 cal=11.2 kcal

Conclusion:

The additional heat required to completely melt ice is 11.2 kcal.

(c)

To determine

The heat required to raise the temperature of water to 27°C.

(c)

Expert Solution

Answer to Problem 3SP

The heat required to raise the temperature of water to 27°C is 3780 cal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

Here,

Q is the heat

m is the mass

c is the specific heat capacity of ice

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Here,

ΔT is the change in temperature

T1 is the initial temperature

T2 is the final temperature

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 140 g for m, 1 cal/g⋅°C for c, 0°C for T1 and 27°C for T2 in equation (3)

Q=140 g×1 cal/g⋅°C(27°C−0°C) =3780 cal

Conclusion:

The heat required to raise the temperature of water to 27°C is 3780 cal.

(d)

To determine

The total heat required to convert ice at −22°C to water at 27 ° C.

(d)

Expert Solution

Answer to Problem 3SP

The total heat required to convert ice at −22°C to water at 27 ° C is 16520 cal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

The total heat will be the sum of the heat required to rise the temperature of ice from −22°C to 0°C, and heat required to completely melt ice at 0°C and the heat required to rise the temperature water obtained by melting ice to 27°C

Qtotal=(mc(T1−T2))ice+mLf+(mc(T1−T2))water

Substitute 1540 cal for the first term, 11200 cal for the second term, 3780 cal for the third term

Qtotal=1540 cal+11200 cal+3780 cal=16520 cal=16.52 kcal

Conclusion:

The total heat required to convert ice at −22°C to water at 27 ° C is 16520 cal.

(e)

To determine

Which takes more energy in the processes mentioned in part a, b, c.

(e)

Expert Solution

Answer to Problem 3SP

Melting of ice completely takes high energy of 11.2 kcal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

The heat required to raise the temperature from −22°C to 0°C is 1540 cal, heat required to melt ice completely is 11.2 kcal and heat required to heat water to 27°C is 3780 cal.

In which more heat is consumed for the intermediate process, which is melting of ice.

Conclusion:

Melting of ice completely takes high energy of 11.2 kcal.

(f)

To determine

To explain whether we can find this total heat simply by computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 140 g water by 49°C .

(f)

Expert Solution

Answer to Problem 3SP

No, before melting the temperature of ice should from −22°C to 0°C. it is also needed to include to complete the calculation.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Three processes are takes place in the conversion of ice to water. Fist ice should be raised from its initial temperature to 0°C and then to completely melt ice to convert to water finally to raise the temperature of water from 0°C to 27°C.

In different stages, the heat capacity will be different. In the intermediate stage during phase change it is required to consider the latent heat of fusion.

Conclusion:

No, before melting the temperature of ice should from −22°C to 0°C. it is also needed to include to complete the calculation.

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Answer 2

Question

Chapter 10, Problem 3SP

(a)

To determine

The heat required to raise the ice to 0°C.

(a)

Expert Solution

Answer to Problem 3SP

The heat required to raise the ice to 0°C is 1540 cal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

Here,

Q is the heat

m is the mass

c is the specific heat capacity of ice

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Here,

ΔT is the change in temperature

T1 is the initial temperature

T2 is the final temperature

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 140 g for m, 0.5 cal/g⋅°C for c, −22°C for T1 and 0°C for T2 in equation (3)

Q=140 g×0.5 cal/g⋅°C(0°C−(−22°C)) =1540 cal

Conclusion:

The heat required to raise the ice to 0°C is 1540 cal.

(b)

To determine

The additional heat required to completely melt ice.

(b)

Expert Solution

Answer to Problem 3SP

The additional heat required to completely melt ice is 11.2 kcal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of latent heat capacity

Q=mLf (1)

Here,

Q is the heat required tot completely melt ice

m is the mass

Lf is the latent heat capacity

Substitute 140 g for m and 80 cal/g for Lf

Q=140 g×80 cal/g=11200 cal=11.2 kcal

Conclusion:

The additional heat required to completely melt ice is 11.2 kcal.

(c)

To determine

The heat required to raise the temperature of water to 27°C.

(c)

Expert Solution

Answer to Problem 3SP

The heat required to raise the temperature of water to 27°C is 3780 cal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

Here,

Q is the heat

m is the mass

c is the specific heat capacity of ice

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Here,

ΔT is the change in temperature

T1 is the initial temperature

T2 is the final temperature

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 140 g for m, 1 cal/g⋅°C for c, 0°C for T1 and 27°C for T2 in equation (3)

Q=140 g×1 cal/g⋅°C(27°C−0°C) =3780 cal

Conclusion:

The heat required to raise the temperature of water to 27°C is 3780 cal.

(d)

To determine

The total heat required to convert ice at −22°C to water at 27 ° C.

(d)

Expert Solution

Answer to Problem 3SP

The total heat required to convert ice at −22°C to water at 27 ° C is 16520 cal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

The total heat will be the sum of the heat required to rise the temperature of ice from −22°C to 0°C, and heat required to completely melt ice at 0°C and the heat required to rise the temperature water obtained by melting ice to 27°C

Qtotal=(mc(T1−T2))ice+mLf+(mc(T1−T2))water

Substitute 1540 cal for the first term, 11200 cal for the second term, 3780 cal for the third term

Qtotal=1540 cal+11200 cal+3780 cal=16520 cal=16.52 kcal

Conclusion:

The total heat required to convert ice at −22°C to water at 27 ° C is 16520 cal.

(e)

To determine

Which takes more energy in the processes mentioned in part a, b, c.

(e)

Expert Solution

Answer to Problem 3SP

Melting of ice completely takes high energy of 11.2 kcal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

The heat required to raise the temperature from −22°C to 0°C is 1540 cal, heat required to melt ice completely is 11.2 kcal and heat required to heat water to 27°C is 3780 cal.

In which more heat is consumed for the intermediate process, which is melting of ice.

Conclusion:

Melting of ice completely takes high energy of 11.2 kcal.

(f)

To determine

To explain whether we can find this total heat simply by computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 140 g water by 49°C .

(f)

Expert Solution

Answer to Problem 3SP

No, before melting the temperature of ice should from −22°C to 0°C. it is also needed to include to complete the calculation.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Three processes are takes place in the conversion of ice to water. Fist ice should be raised from its initial temperature to 0°C and then to completely melt ice to convert to water finally to raise the temperature of water from 0°C to 27°C.

In different stages, the heat capacity will be different. In the intermediate stage during phase change it is required to consider the latent heat of fusion.

Conclusion:

No, before melting the temperature of ice should from −22°C to 0°C. it is also needed to include to complete the calculation.

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Answer 3

Question

Chapter 10, Problem 3SP

(a)

To determine

The heat required to raise the ice to 0°C.

(a)

Expert Solution

Answer to Problem 3SP

The heat required to raise the ice to 0°C is 1540 cal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

Here,

Q is the heat

m is the mass

c is the specific heat capacity of ice

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Here,

ΔT is the change in temperature

T1 is the initial temperature

T2 is the final temperature

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 140 g for m, 0.5 cal/g⋅°C for c, −22°C for T1 and 0°C for T2 in equation (3)

Q=140 g×0.5 cal/g⋅°C(0°C−(−22°C)) =1540 cal

Conclusion:

The heat required to raise the ice to 0°C is 1540 cal.

(b)

To determine

The additional heat required to completely melt ice.

(b)

Expert Solution

Answer to Problem 3SP

The additional heat required to completely melt ice is 11.2 kcal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of latent heat capacity

Q=mLf (1)

Here,

Q is the heat required tot completely melt ice

m is the mass

Lf is the latent heat capacity

Substitute 140 g for m and 80 cal/g for Lf

Q=140 g×80 cal/g=11200 cal=11.2 kcal

Conclusion:

The additional heat required to completely melt ice is 11.2 kcal.

(c)

To determine

The heat required to raise the temperature of water to 27°C.

(c)

Expert Solution

Answer to Problem 3SP

The heat required to raise the temperature of water to 27°C is 3780 cal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

Here,

Q is the heat

m is the mass

c is the specific heat capacity of ice

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Here,

ΔT is the change in temperature

T1 is the initial temperature

T2 is the final temperature

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 140 g for m, 1 cal/g⋅°C for c, 0°C for T1 and 27°C for T2 in equation (3)

Q=140 g×1 cal/g⋅°C(27°C−0°C) =3780 cal

Conclusion:

The heat required to raise the temperature of water to 27°C is 3780 cal.

(d)

To determine

The total heat required to convert ice at −22°C to water at 27 ° C.

(d)

Expert Solution

Answer to Problem 3SP

The total heat required to convert ice at −22°C to water at 27 ° C is 16520 cal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

The total heat will be the sum of the heat required to rise the temperature of ice from −22°C to 0°C, and heat required to completely melt ice at 0°C and the heat required to rise the temperature water obtained by melting ice to 27°C

Qtotal=(mc(T1−T2))ice+mLf+(mc(T1−T2))water

Substitute 1540 cal for the first term, 11200 cal for the second term, 3780 cal for the third term

Qtotal=1540 cal+11200 cal+3780 cal=16520 cal=16.52 kcal

Conclusion:

The total heat required to convert ice at −22°C to water at 27 ° C is 16520 cal.

(e)

To determine

Which takes more energy in the processes mentioned in part a, b, c.

(e)

Expert Solution

Answer to Problem 3SP

Melting of ice completely takes high energy of 11.2 kcal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

The heat required to raise the temperature from −22°C to 0°C is 1540 cal, heat required to melt ice completely is 11.2 kcal and heat required to heat water to 27°C is 3780 cal.

In which more heat is consumed for the intermediate process, which is melting of ice.

Conclusion:

Melting of ice completely takes high energy of 11.2 kcal.

(f)

To determine

To explain whether we can find this total heat simply by computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 140 g water by 49°C .

(f)

Expert Solution

Answer to Problem 3SP

No, before melting the temperature of ice should from −22°C to 0°C. it is also needed to include to complete the calculation.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Three processes are takes place in the conversion of ice to water. Fist ice should be raised from its initial temperature to 0°C and then to completely melt ice to convert to water finally to raise the temperature of water from 0°C to 27°C.

In different stages, the heat capacity will be different. In the intermediate stage during phase change it is required to consider the latent heat of fusion.

Conclusion:

No, before melting the temperature of ice should from −22°C to 0°C. it is also needed to include to complete the calculation.

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Answer 4

Question

Chapter 10, Problem 3SP

(a)

To determine

The heat required to raise the ice to 0°C.

(a)

Expert Solution

Answer to Problem 3SP

The heat required to raise the ice to 0°C is 1540 cal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

Here,

Q is the heat

m is the mass

c is the specific heat capacity of ice

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Here,

ΔT is the change in temperature

T1 is the initial temperature

T2 is the final temperature

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 140 g for m, 0.5 cal/g⋅°C for c, −22°C for T1 and 0°C for T2 in equation (3)

Q=140 g×0.5 cal/g⋅°C(0°C−(−22°C)) =1540 cal

Conclusion:

The heat required to raise the ice to 0°C is 1540 cal.

(b)

To determine

The additional heat required to completely melt ice.

(b)

Expert Solution

Answer to Problem 3SP

The additional heat required to completely melt ice is 11.2 kcal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of latent heat capacity

Q=mLf (1)

Here,

Q is the heat required tot completely melt ice

m is the mass

Lf is the latent heat capacity

Substitute 140 g for m and 80 cal/g for Lf

Q=140 g×80 cal/g=11200 cal=11.2 kcal

Conclusion:

The additional heat required to completely melt ice is 11.2 kcal.

(c)

To determine

The heat required to raise the temperature of water to 27°C.

(c)

Expert Solution

Answer to Problem 3SP

The heat required to raise the temperature of water to 27°C is 3780 cal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

Here,

Q is the heat

m is the mass

c is the specific heat capacity of ice

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Here,

ΔT is the change in temperature

T1 is the initial temperature

T2 is the final temperature

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 140 g for m, 1 cal/g⋅°C for c, 0°C for T1 and 27°C for T2 in equation (3)

Q=140 g×1 cal/g⋅°C(27°C−0°C) =3780 cal

Conclusion:

The heat required to raise the temperature of water to 27°C is 3780 cal.

(d)

To determine

The total heat required to convert ice at −22°C to water at 27 ° C.

(d)

Expert Solution

Answer to Problem 3SP

The total heat required to convert ice at −22°C to water at 27 ° C is 16520 cal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

The total heat will be the sum of the heat required to rise the temperature of ice from −22°C to 0°C, and heat required to completely melt ice at 0°C and the heat required to rise the temperature water obtained by melting ice to 27°C

Qtotal=(mc(T1−T2))ice+mLf+(mc(T1−T2))water

Substitute 1540 cal for the first term, 11200 cal for the second term, 3780 cal for the third term

Qtotal=1540 cal+11200 cal+3780 cal=16520 cal=16.52 kcal

Conclusion:

The total heat required to convert ice at −22°C to water at 27 ° C is 16520 cal.

(e)

To determine

Which takes more energy in the processes mentioned in part a, b, c.

(e)

Expert Solution

Answer to Problem 3SP

Melting of ice completely takes high energy of 11.2 kcal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

The heat required to raise the temperature from −22°C to 0°C is 1540 cal, heat required to melt ice completely is 11.2 kcal and heat required to heat water to 27°C is 3780 cal.

In which more heat is consumed for the intermediate process, which is melting of ice.

Conclusion:

Melting of ice completely takes high energy of 11.2 kcal.

(f)

To determine

To explain whether we can find this total heat simply by computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 140 g water by 49°C .

(f)

Expert Solution

Answer to Problem 3SP

No, before melting the temperature of ice should from −22°C to 0°C. it is also needed to include to complete the calculation.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Three processes are takes place in the conversion of ice to water. Fist ice should be raised from its initial temperature to 0°C and then to completely melt ice to convert to water finally to raise the temperature of water from 0°C to 27°C.

In different stages, the heat capacity will be different. In the intermediate stage during phase change it is required to consider the latent heat of fusion.

Conclusion:

No, before melting the temperature of ice should from −22°C to 0°C. it is also needed to include to complete the calculation.

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Answer 5

Chapter 10, Problem 3SP

(a)

To determine

The heat required to raise the ice to 0°C.

(a)

Expert Solution

Answer to Problem 3SP

The heat required to raise the ice to 0°C is 1540 cal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

Here,

Q is the heat

m is the mass

c is the specific heat capacity of ice

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Here,

ΔT is the change in temperature

T1 is the initial temperature

T2 is the final temperature

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 140 g for m, 0.5 cal/g⋅°C for c, −22°C for T1 and 0°C for T2 in equation (3)

Q=140 g×0.5 cal/g⋅°C(0°C−(−22°C)) =1540 cal

Conclusion:

The heat required to raise the ice to 0°C is 1540 cal.

(b)

To determine

The additional heat required to completely melt ice.

(b)

Expert Solution

Answer to Problem 3SP

The additional heat required to completely melt ice is 11.2 kcal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of latent heat capacity

Q=mLf (1)

Here,

Q is the heat required tot completely melt ice

m is the mass

Lf is the latent heat capacity

Substitute 140 g for m and 80 cal/g for Lf

Q=140 g×80 cal/g=11200 cal=11.2 kcal

Conclusion:

The additional heat required to completely melt ice is 11.2 kcal.

(c)

To determine

The heat required to raise the temperature of water to 27°C.

(c)

Expert Solution

Answer to Problem 3SP

The heat required to raise the temperature of water to 27°C is 3780 cal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

Here,

Q is the heat

m is the mass

c is the specific heat capacity of ice

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Here,

ΔT is the change in temperature

T1 is the initial temperature

T2 is the final temperature

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 140 g for m, 1 cal/g⋅°C for c, 0°C for T1 and 27°C for T2 in equation (3)

Q=140 g×1 cal/g⋅°C(27°C−0°C) =3780 cal

Conclusion:

The heat required to raise the temperature of water to 27°C is 3780 cal.

(d)

To determine

The total heat required to convert ice at −22°C to water at 27 ° C.

(d)

Expert Solution

Answer to Problem 3SP

The total heat required to convert ice at −22°C to water at 27 ° C is 16520 cal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

The total heat will be the sum of the heat required to rise the temperature of ice from −22°C to 0°C, and heat required to completely melt ice at 0°C and the heat required to rise the temperature water obtained by melting ice to 27°C

Qtotal=(mc(T1−T2))ice+mLf+(mc(T1−T2))water

Substitute 1540 cal for the first term, 11200 cal for the second term, 3780 cal for the third term

Qtotal=1540 cal+11200 cal+3780 cal=16520 cal=16.52 kcal

Conclusion:

The total heat required to convert ice at −22°C to water at 27 ° C is 16520 cal.

(e)

To determine

Which takes more energy in the processes mentioned in part a, b, c.

(e)

Expert Solution

Answer to Problem 3SP

Melting of ice completely takes high energy of 11.2 kcal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

The heat required to raise the temperature from −22°C to 0°C is 1540 cal, heat required to melt ice completely is 11.2 kcal and heat required to heat water to 27°C is 3780 cal.

In which more heat is consumed for the intermediate process, which is melting of ice.

Conclusion:

Melting of ice completely takes high energy of 11.2 kcal.

(f)

To determine

To explain whether we can find this total heat simply by computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 140 g water by 49°C .

(f)

Expert Solution

Answer to Problem 3SP

No, before melting the temperature of ice should from −22°C to 0°C. it is also needed to include to complete the calculation.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Three processes are takes place in the conversion of ice to water. Fist ice should be raised from its initial temperature to 0°C and then to completely melt ice to convert to water finally to raise the temperature of water from 0°C to 27°C.

In different stages, the heat capacity will be different. In the intermediate stage during phase change it is required to consider the latent heat of fusion.

Conclusion:

No, before melting the temperature of ice should from −22°C to 0°C. it is also needed to include to complete the calculation.

Answer 6

Chapter 10, Problem 3SP

(a)

To determine

The heat required to raise the ice to 0°C.

(a)

Expert Solution

Answer to Problem 3SP

The heat required to raise the ice to 0°C is 1540 cal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

Here,

Q is the heat

m is the mass

c is the specific heat capacity of ice

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Here,

ΔT is the change in temperature

T1 is the initial temperature

T2 is the final temperature

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 140 g for m, 0.5 cal/g⋅°C for c, −22°C for T1 and 0°C for T2 in equation (3)

Q=140 g×0.5 cal/g⋅°C(0°C−(−22°C)) =1540 cal

Conclusion:

The heat required to raise the ice to 0°C is 1540 cal.

Answer 7

Chapter 10, Problem 3SP

(a)

To determine

The heat required to raise the ice to 0°C.

Answer 8

(a)

Expert Solution

Answer to Problem 3SP

The heat required to raise the ice to 0°C is 1540 cal.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

Here,

Q is the heat

m is the mass

c is the specific heat capacity of ice

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Here,

ΔT is the change in temperature

T1 is the initial temperature

T2 is the final temperature

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 140 g for m, 0.5 cal/g⋅°C for c, −22°C for T1 and 0°C for T2 in equation (3)

Q=140 g×0.5 cal/g⋅°C(0°C−(−22°C)) =1540 cal

Conclusion:

The heat required to raise the ice to 0°C is 1540 cal.

Answer 13

(b)

To determine

The additional heat required to completely melt ice.

(b)

Expert Solution

Answer to Problem 3SP

The additional heat required to completely melt ice is 11.2 kcal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of latent heat capacity

Q=mLf (1)

Here,

Q is the heat required tot completely melt ice

m is the mass

Lf is the latent heat capacity

Substitute 140 g for m and 80 cal/g for Lf

Q=140 g×80 cal/g=11200 cal=11.2 kcal

Conclusion:

The additional heat required to completely melt ice is 11.2 kcal.

Answer 14

(b)

To determine

The additional heat required to completely melt ice.

Answer 15

(b)

Expert Solution

Answer to Problem 3SP

The additional heat required to completely melt ice is 11.2 kcal.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of latent heat capacity

Q=mLf (1)

Here,

Q is the heat required tot completely melt ice

m is the mass

Lf is the latent heat capacity

Substitute 140 g for m and 80 cal/g for Lf

Q=140 g×80 cal/g=11200 cal=11.2 kcal

Conclusion:

The additional heat required to completely melt ice is 11.2 kcal.

Answer 20

(c)

To determine

The heat required to raise the temperature of water to 27°C.

(c)

Expert Solution

Answer to Problem 3SP

The heat required to raise the temperature of water to 27°C is 3780 cal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

Here,

Q is the heat

m is the mass

c is the specific heat capacity of ice

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Here,

ΔT is the change in temperature

T1 is the initial temperature

T2 is the final temperature

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 140 g for m, 1 cal/g⋅°C for c, 0°C for T1 and 27°C for T2 in equation (3)

Q=140 g×1 cal/g⋅°C(27°C−0°C) =3780 cal

Conclusion:

The heat required to raise the temperature of water to 27°C is 3780 cal.

Answer 21

(c)

To determine

The heat required to raise the temperature of water to 27°C.

Answer 22

(c)

Expert Solution

Answer to Problem 3SP

The heat required to raise the temperature of water to 27°C is 3780 cal.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

Here,

Q is the heat

m is the mass

c is the specific heat capacity of ice

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Here,

ΔT is the change in temperature

T1 is the initial temperature

T2 is the final temperature

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 140 g for m, 1 cal/g⋅°C for c, 0°C for T1 and 27°C for T2 in equation (3)

Q=140 g×1 cal/g⋅°C(27°C−0°C) =3780 cal

Conclusion:

The heat required to raise the temperature of water to 27°C is 3780 cal.

Answer 27

(d)

To determine

The total heat required to convert ice at −22°C to water at 27 ° C.

(d)

Expert Solution

Answer to Problem 3SP

The total heat required to convert ice at −22°C to water at 27 ° C is 16520 cal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

The total heat will be the sum of the heat required to rise the temperature of ice from −22°C to 0°C, and heat required to completely melt ice at 0°C and the heat required to rise the temperature water obtained by melting ice to 27°C

Qtotal=(mc(T1−T2))ice+mLf+(mc(T1−T2))water

Substitute 1540 cal for the first term, 11200 cal for the second term, 3780 cal for the third term

Qtotal=1540 cal+11200 cal+3780 cal=16520 cal=16.52 kcal

Conclusion:

The total heat required to convert ice at −22°C to water at 27 ° C is 16520 cal.

Answer 28

(d)

To determine

The total heat required to convert ice at −22°C to water at 27 ° C.

Answer 29

(d)

Expert Solution

Answer to Problem 3SP

The total heat required to convert ice at −22°C to water at 27 ° C is 16520 cal.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

The total heat will be the sum of the heat required to rise the temperature of ice from −22°C to 0°C, and heat required to completely melt ice at 0°C and the heat required to rise the temperature water obtained by melting ice to 27°C

Qtotal=(mc(T1−T2))ice+mLf+(mc(T1−T2))water

Substitute 1540 cal for the first term, 11200 cal for the second term, 3780 cal for the third term

Qtotal=1540 cal+11200 cal+3780 cal=16520 cal=16.52 kcal

Conclusion:

The total heat required to convert ice at −22°C to water at 27 ° C is 16520 cal.

Answer 34

(e)

To determine

Which takes more energy in the processes mentioned in part a, b, c.

(e)

Expert Solution

Answer to Problem 3SP

Melting of ice completely takes high energy of 11.2 kcal.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

The heat required to raise the temperature from −22°C to 0°C is 1540 cal, heat required to melt ice completely is 11.2 kcal and heat required to heat water to 27°C is 3780 cal.

In which more heat is consumed for the intermediate process, which is melting of ice.

Conclusion:

Melting of ice completely takes high energy of 11.2 kcal.

Answer 35

(e)

To determine

Which takes more energy in the processes mentioned in part a, b, c.

Answer 36

(e)

Expert Solution

Answer to Problem 3SP

Melting of ice completely takes high energy of 11.2 kcal.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

The heat required to raise the temperature from −22°C to 0°C is 1540 cal, heat required to melt ice completely is 11.2 kcal and heat required to heat water to 27°C is 3780 cal.

In which more heat is consumed for the intermediate process, which is melting of ice.

Conclusion:

Melting of ice completely takes high energy of 11.2 kcal.

Answer 41

(f)

To determine

To explain whether we can find this total heat simply by computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 140 g water by 49°C .

(f)

Expert Solution

Answer to Problem 3SP

No, before melting the temperature of ice should from −22°C to 0°C. it is also needed to include to complete the calculation.

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Three processes are takes place in the conversion of ice to water. Fist ice should be raised from its initial temperature to 0°C and then to completely melt ice to convert to water finally to raise the temperature of water from 0°C to 27°C.

In different stages, the heat capacity will be different. In the intermediate stage during phase change it is required to consider the latent heat of fusion.

Conclusion:

No, before melting the temperature of ice should from −22°C to 0°C. it is also needed to include to complete the calculation.

Answer 42

(f)

To determine

To explain whether we can find this total heat simply by computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 140 g water by 49°C .

Answer 43

(f)

Expert Solution

Answer to Problem 3SP

No, before melting the temperature of ice should from −22°C to 0°C. it is also needed to include to complete the calculation.

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

Given info: The mass of ice is 140 g at initial temperature −22°C. The specific heat capacity of ice is 0.5 cal/g⋅°C and the specific heat capacity of water is 1 cal/g⋅°C.

Three processes are takes place in the conversion of ice to water. Fist ice should be raised from its initial temperature to 0°C and then to completely melt ice to convert to water finally to raise the temperature of water from 0°C to 27°C.

In different stages, the heat capacity will be different. In the intermediate stage during phase change it is required to consider the latent heat of fusion.

Conclusion:

No, before melting the temperature of ice should from −22°C to 0°C. it is also needed to include to complete the calculation.