The amount of heat transferred from metal to water.

Question

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Answer 1

Question

Chapter 10, Problem 4SP

(a)

To determine

The amount of heat transferred from metal to water.

(a)

Expert Solution

Answer to Problem 4SP

The amount of heat transferred from metal to water is 2340 cal.

Explanation of Solution

Given info: A 135 g quantity of a certain metal, initially at 130°C , is dropped into an insulated beaker containing 90 g of water at 20°C . The final temperature of the metal and water in the beaker is 46°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 90 g for m, 1 cal/g⋅°C for c, 20°C for T1 and 46°C for T2 in equation (3)

Q=90 g×1 cal/g⋅°C(46°C−20°C) =2340 cal

Conclusion:

The amount of heat transferred from metal to water is 2340 cal.

(b)

To determine

The specific heat capacity of metal.

(b)

Expert Solution

Answer to Problem 4SP

The specific heat capacity of metal is 0.206 cal/g⋅°C.

Explanation of Solution

Write the expression for specific heat capacity of metal

c=Qm(T2−T1)

Substitute 135 g for m, 2340 cal for Q, 130°C for T1, and 46°C for T2 in the above equation

c=2340 cal135 g(130°C−46°C)=0.206 cal/g⋅°C

Conclusion:

The specific heat capacity of metal is 0.206 cal/g⋅°C.

(c)

To determine

The quantity of metal dropped into the insulating beaker.

(c)

Expert Solution

Answer to Problem 4SP

The quantity of metal dropped into the insulating beaker is 360 g.

Explanation of Solution

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 90 g for m, 1 cal/g⋅°C for c, 20°C for T1 and 70°C for T2 in equation (3)

Q=90 g×1 cal/g⋅°C(70°C−20°C) =4500 cal

Write the expression of mass in terms of specific heat capacity

m=QcΔT (4)

Substitute 0.206 cal/g⋅°C for c, 4500 cal for Q, 130°C for T1 and 70°C for T2 in equation (4)

m=4500 cal0.206 cal/g⋅°C×(130°C−70°C)=364 g≈360 g

The negative sign in the above equation is avoided since mass is always a positive quantity.

Conclusion:

The quantity of heat was dropped from metal to insulating beaker is 360 g.

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Answer 2

Question

Chapter 10, Problem 4SP

(a)

To determine

The amount of heat transferred from metal to water.

(a)

Expert Solution

Answer to Problem 4SP

The amount of heat transferred from metal to water is 2340 cal.

Explanation of Solution

Given info: A 135 g quantity of a certain metal, initially at 130°C , is dropped into an insulated beaker containing 90 g of water at 20°C . The final temperature of the metal and water in the beaker is 46°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 90 g for m, 1 cal/g⋅°C for c, 20°C for T1 and 46°C for T2 in equation (3)

Q=90 g×1 cal/g⋅°C(46°C−20°C) =2340 cal

Conclusion:

The amount of heat transferred from metal to water is 2340 cal.

(b)

To determine

The specific heat capacity of metal.

(b)

Expert Solution

Answer to Problem 4SP

The specific heat capacity of metal is 0.206 cal/g⋅°C.

Explanation of Solution

Write the expression for specific heat capacity of metal

c=Qm(T2−T1)

Substitute 135 g for m, 2340 cal for Q, 130°C for T1, and 46°C for T2 in the above equation

c=2340 cal135 g(130°C−46°C)=0.206 cal/g⋅°C

Conclusion:

The specific heat capacity of metal is 0.206 cal/g⋅°C.

(c)

To determine

The quantity of metal dropped into the insulating beaker.

(c)

Expert Solution

Answer to Problem 4SP

The quantity of metal dropped into the insulating beaker is 360 g.

Explanation of Solution

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 90 g for m, 1 cal/g⋅°C for c, 20°C for T1 and 70°C for T2 in equation (3)

Q=90 g×1 cal/g⋅°C(70°C−20°C) =4500 cal

Write the expression of mass in terms of specific heat capacity

m=QcΔT (4)

Substitute 0.206 cal/g⋅°C for c, 4500 cal for Q, 130°C for T1 and 70°C for T2 in equation (4)

m=4500 cal0.206 cal/g⋅°C×(130°C−70°C)=364 g≈360 g

The negative sign in the above equation is avoided since mass is always a positive quantity.

Conclusion:

The quantity of heat was dropped from metal to insulating beaker is 360 g.

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Answer 3

Question

Chapter 10, Problem 4SP

(a)

To determine

The amount of heat transferred from metal to water.

(a)

Expert Solution

Answer to Problem 4SP

The amount of heat transferred from metal to water is 2340 cal.

Explanation of Solution

Given info: A 135 g quantity of a certain metal, initially at 130°C , is dropped into an insulated beaker containing 90 g of water at 20°C . The final temperature of the metal and water in the beaker is 46°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 90 g for m, 1 cal/g⋅°C for c, 20°C for T1 and 46°C for T2 in equation (3)

Q=90 g×1 cal/g⋅°C(46°C−20°C) =2340 cal

Conclusion:

The amount of heat transferred from metal to water is 2340 cal.

(b)

To determine

The specific heat capacity of metal.

(b)

Expert Solution

Answer to Problem 4SP

The specific heat capacity of metal is 0.206 cal/g⋅°C.

Explanation of Solution

Write the expression for specific heat capacity of metal

c=Qm(T2−T1)

Substitute 135 g for m, 2340 cal for Q, 130°C for T1, and 46°C for T2 in the above equation

c=2340 cal135 g(130°C−46°C)=0.206 cal/g⋅°C

Conclusion:

The specific heat capacity of metal is 0.206 cal/g⋅°C.

(c)

To determine

The quantity of metal dropped into the insulating beaker.

(c)

Expert Solution

Answer to Problem 4SP

The quantity of metal dropped into the insulating beaker is 360 g.

Explanation of Solution

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 90 g for m, 1 cal/g⋅°C for c, 20°C for T1 and 70°C for T2 in equation (3)

Q=90 g×1 cal/g⋅°C(70°C−20°C) =4500 cal

Write the expression of mass in terms of specific heat capacity

m=QcΔT (4)

Substitute 0.206 cal/g⋅°C for c, 4500 cal for Q, 130°C for T1 and 70°C for T2 in equation (4)

m=4500 cal0.206 cal/g⋅°C×(130°C−70°C)=364 g≈360 g

The negative sign in the above equation is avoided since mass is always a positive quantity.

Conclusion:

The quantity of heat was dropped from metal to insulating beaker is 360 g.

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Answer 4

Question

Chapter 10, Problem 4SP

(a)

To determine

The amount of heat transferred from metal to water.

(a)

Expert Solution

Answer to Problem 4SP

The amount of heat transferred from metal to water is 2340 cal.

Explanation of Solution

Given info: A 135 g quantity of a certain metal, initially at 130°C , is dropped into an insulated beaker containing 90 g of water at 20°C . The final temperature of the metal and water in the beaker is 46°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 90 g for m, 1 cal/g⋅°C for c, 20°C for T1 and 46°C for T2 in equation (3)

Q=90 g×1 cal/g⋅°C(46°C−20°C) =2340 cal

Conclusion:

The amount of heat transferred from metal to water is 2340 cal.

(b)

To determine

The specific heat capacity of metal.

(b)

Expert Solution

Answer to Problem 4SP

The specific heat capacity of metal is 0.206 cal/g⋅°C.

Explanation of Solution

Write the expression for specific heat capacity of metal

c=Qm(T2−T1)

Substitute 135 g for m, 2340 cal for Q, 130°C for T1, and 46°C for T2 in the above equation

c=2340 cal135 g(130°C−46°C)=0.206 cal/g⋅°C

Conclusion:

The specific heat capacity of metal is 0.206 cal/g⋅°C.

(c)

To determine

The quantity of metal dropped into the insulating beaker.

(c)

Expert Solution

Answer to Problem 4SP

The quantity of metal dropped into the insulating beaker is 360 g.

Explanation of Solution

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 90 g for m, 1 cal/g⋅°C for c, 20°C for T1 and 70°C for T2 in equation (3)

Q=90 g×1 cal/g⋅°C(70°C−20°C) =4500 cal

Write the expression of mass in terms of specific heat capacity

m=QcΔT (4)

Substitute 0.206 cal/g⋅°C for c, 4500 cal for Q, 130°C for T1 and 70°C for T2 in equation (4)

m=4500 cal0.206 cal/g⋅°C×(130°C−70°C)=364 g≈360 g

The negative sign in the above equation is avoided since mass is always a positive quantity.

Conclusion:

The quantity of heat was dropped from metal to insulating beaker is 360 g.

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Answer 5

Chapter 10, Problem 4SP

(a)

To determine

The amount of heat transferred from metal to water.

(a)

Expert Solution

Answer to Problem 4SP

The amount of heat transferred from metal to water is 2340 cal.

Explanation of Solution

Given info: A 135 g quantity of a certain metal, initially at 130°C , is dropped into an insulated beaker containing 90 g of water at 20°C . The final temperature of the metal and water in the beaker is 46°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 90 g for m, 1 cal/g⋅°C for c, 20°C for T1 and 46°C for T2 in equation (3)

Q=90 g×1 cal/g⋅°C(46°C−20°C) =2340 cal

Conclusion:

The amount of heat transferred from metal to water is 2340 cal.

(b)

To determine

The specific heat capacity of metal.

(b)

Expert Solution

Answer to Problem 4SP

The specific heat capacity of metal is 0.206 cal/g⋅°C.

Explanation of Solution

Write the expression for specific heat capacity of metal

c=Qm(T2−T1)

Substitute 135 g for m, 2340 cal for Q, 130°C for T1, and 46°C for T2 in the above equation

c=2340 cal135 g(130°C−46°C)=0.206 cal/g⋅°C

Conclusion:

The specific heat capacity of metal is 0.206 cal/g⋅°C.

(c)

To determine

The quantity of metal dropped into the insulating beaker.

(c)

Expert Solution

Answer to Problem 4SP

The quantity of metal dropped into the insulating beaker is 360 g.

Explanation of Solution

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 90 g for m, 1 cal/g⋅°C for c, 20°C for T1 and 70°C for T2 in equation (3)

Q=90 g×1 cal/g⋅°C(70°C−20°C) =4500 cal

Write the expression of mass in terms of specific heat capacity

m=QcΔT (4)

Substitute 0.206 cal/g⋅°C for c, 4500 cal for Q, 130°C for T1 and 70°C for T2 in equation (4)

m=4500 cal0.206 cal/g⋅°C×(130°C−70°C)=364 g≈360 g

The negative sign in the above equation is avoided since mass is always a positive quantity.

Conclusion:

The quantity of heat was dropped from metal to insulating beaker is 360 g.

Answer 6

Chapter 10, Problem 4SP

(a)

To determine

The amount of heat transferred from metal to water.

(a)

Expert Solution

Answer to Problem 4SP

The amount of heat transferred from metal to water is 2340 cal.

Explanation of Solution

Given info: A 135 g quantity of a certain metal, initially at 130°C , is dropped into an insulated beaker containing 90 g of water at 20°C . The final temperature of the metal and water in the beaker is 46°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 90 g for m, 1 cal/g⋅°C for c, 20°C for T1 and 46°C for T2 in equation (3)

Q=90 g×1 cal/g⋅°C(46°C−20°C) =2340 cal

Conclusion:

The amount of heat transferred from metal to water is 2340 cal.

Answer 7

Chapter 10, Problem 4SP

(a)

To determine

The amount of heat transferred from metal to water.

Answer 8

(a)

Expert Solution

Answer to Problem 4SP

The amount of heat transferred from metal to water is 2340 cal.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given info: A 135 g quantity of a certain metal, initially at 130°C , is dropped into an insulated beaker containing 90 g of water at 20°C . The final temperature of the metal and water in the beaker is 46°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 90 g for m, 1 cal/g⋅°C for c, 20°C for T1 and 46°C for T2 in equation (3)

Q=90 g×1 cal/g⋅°C(46°C−20°C) =2340 cal

Conclusion:

The amount of heat transferred from metal to water is 2340 cal.

Answer 13

(b)

To determine

The specific heat capacity of metal.

(b)

Expert Solution

Answer to Problem 4SP

The specific heat capacity of metal is 0.206 cal/g⋅°C.

Explanation of Solution

Write the expression for specific heat capacity of metal

c=Qm(T2−T1)

Substitute 135 g for m, 2340 cal for Q, 130°C for T1, and 46°C for T2 in the above equation

c=2340 cal135 g(130°C−46°C)=0.206 cal/g⋅°C

Conclusion:

The specific heat capacity of metal is 0.206 cal/g⋅°C.

Answer 14

(b)

To determine

The specific heat capacity of metal.

Answer 15

(b)

Expert Solution

Answer to Problem 4SP

The specific heat capacity of metal is 0.206 cal/g⋅°C.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the expression for specific heat capacity of metal

c=Qm(T2−T1)

Substitute 135 g for m, 2340 cal for Q, 130°C for T1, and 46°C for T2 in the above equation

c=2340 cal135 g(130°C−46°C)=0.206 cal/g⋅°C

Conclusion:

The specific heat capacity of metal is 0.206 cal/g⋅°C.

Answer 20

(c)

To determine

The quantity of metal dropped into the insulating beaker.

(c)

Expert Solution

Answer to Problem 4SP

The quantity of metal dropped into the insulating beaker is 360 g.

Explanation of Solution

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 90 g for m, 1 cal/g⋅°C for c, 20°C for T1 and 70°C for T2 in equation (3)

Q=90 g×1 cal/g⋅°C(70°C−20°C) =4500 cal

Write the expression of mass in terms of specific heat capacity

m=QcΔT (4)

Substitute 0.206 cal/g⋅°C for c, 4500 cal for Q, 130°C for T1 and 70°C for T2 in equation (4)

m=4500 cal0.206 cal/g⋅°C×(130°C−70°C)=364 g≈360 g

The negative sign in the above equation is avoided since mass is always a positive quantity.

Conclusion:

The quantity of heat was dropped from metal to insulating beaker is 360 g.

Answer 21

(c)

To determine

The quantity of metal dropped into the insulating beaker.

Answer 22

(c)

Expert Solution

Answer to Problem 4SP

The quantity of metal dropped into the insulating beaker is 360 g.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write the expression for heat in terms of specific heat capacity

Q=mcΔT (1)

ΔT is the change in temperature

Write the expression for change in temperature

ΔT = T2−T1 (2)

Substitute equation (2) in (1)

Q=mc(T2−T1) (3)

Substitute 90 g for m, 1 cal/g⋅°C for c, 20°C for T1 and 70°C for T2 in equation (3)

Q=90 g×1 cal/g⋅°C(70°C−20°C) =4500 cal

Write the expression of mass in terms of specific heat capacity

m=QcΔT (4)

Substitute 0.206 cal/g⋅°C for c, 4500 cal for Q, 130°C for T1 and 70°C for T2 in equation (4)

m=4500 cal0.206 cal/g⋅°C×(130°C−70°C)=364 g≈360 g

The negative sign in the above equation is avoided since mass is always a positive quantity.

Conclusion:

The quantity of heat was dropped from metal to insulating beaker is 360 g.

Answer 27

Ch. 10 - Is an object that has a temperature of 0C hotter...Ch. 10 - Prob. 2CQ Ch. 10 - The volume of a gas held at constant pressure...Ch. 10 - We sometimes attempt to determine whether another...Ch. 10 - Prob. 5CQ Ch. 10 - Is it possible for a temperature to be lower than...Ch. 10 - Is an object with a temperature of 273.2 K hotter...Ch. 10 - Two objects at different temperatures are placed...Ch. 10 - Is it possible for the final temperature of the...Ch. 10 - Two objects of the same mass, but made of...

The amount of heat transferred from metal to water.

The amount of heat transferred from metal to water.

Answer to Problem 4SP

Explanation of Solution

The specific heat capacity of metal.

Answer to Problem 4SP

Explanation of Solution

The quantity of metal dropped into the insulating beaker.

Answer to Problem 4SP

Explanation of Solution

Want to see more full solutions like this?

Chapter 10 Solutions