A 170-g quantity of a certain metal, initially at 125°C, is dropped into an insulated beaker containing 84 g of water at 16°C. The final temperature of the metal and water in the beaker is measured as 39°C. Assume that the heat capacity of the beaker can be ignored. a. How much heat has been transferred from the metal to the water? b. Given the temperature change and mass of the metal, what is the specific heat capacity of the metal? c. If the final temperature of the water and this metal is 53°C instead of 39°C, what quantity of this metal (initially at 125°C) was dropped into the insulated beaker?

Question

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Answer 1

Textbook Question

Chapter 10, Problem 4SP

A 170-g quantity of a certain metal, initially at 125°C, is dropped into an insulated beaker containing 84 g of water at 16°C. The final temperature of the metal and water in the beaker is measured as 39°C. Assume that the heat capacity of the beaker can be ignored.

a. How much heat has been transferred from the metal to the water?

b. Given the temperature change and mass of the metal, what is the specific heat capacity of the metal?

c. If the final temperature of the water and this metal is 53°C instead of 39°C, what quantity of this metal (initially at 125°C) was dropped into the insulated beaker?

(a)

Expert Solution

To determine

The amount of heat transferred from metal to water.

Answer to Problem 4SP

The amount of heat transferred from metal to water is 1932 cal.

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°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 8 4g for m, 1 cal/g⋅°C for c, 16°C for T1 and 39°C for T2 in equation (3)

Q=84 g×1 cal/g⋅°C(39°C−16°C)=1932 cal

Conclusion:

The amount of heat transferred from metal to water is 1932 cal.

(b)

Expert Solution

To determine

The specific heat capacity of metal.

Answer to Problem 4SP

The specific heat capacity of metal is 0.132 cal/g⋅°C.

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°C.

Write the expression for specific heat capacity of metal

c=Qm(T2−T1)

Substitute 170 g for m, 1932 cal for Q, 125°C for T1, and 39°C for T2 in the above equation

c=1932 cal170 g(39°C−125°C)=1932 cal170 g×(86°C)=0.132 cal/g⋅°C

The negative sign in the above expression is avoided since specific heat capacity is a positive quantity.

Conclusion:

The specific heat capacity of metal is 0.132 cal/g⋅°C.

(c)

Expert Solution

To determine

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

Answer to Problem 4SP

The quantity of heat was dropped from metal to insulating beaker is 327 g.

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°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 84 g for m, 1 cal/g⋅°C for c, 16°C for T1 and 53°C for T2 in equation (3)

Q=84 g×1 cal/g⋅°C(53°C−16°C) =3108 cal

Write the expression of mass in terms of specific heat capacity

m=Qc(T2−T1) (4)

Substitute 0.132 cal/g⋅°C for c, 3108 cal for Q, 125°C for T1 and 53°C for T2 in equation (4)

m=3108 cal0.132 cal/g⋅°C×(53°C−125°C)=327 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 327 g.

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Answer 2

Textbook Question

Chapter 10, Problem 4SP

A 170-g quantity of a certain metal, initially at 125°C, is dropped into an insulated beaker containing 84 g of water at 16°C. The final temperature of the metal and water in the beaker is measured as 39°C. Assume that the heat capacity of the beaker can be ignored.

a. How much heat has been transferred from the metal to the water?

b. Given the temperature change and mass of the metal, what is the specific heat capacity of the metal?

c. If the final temperature of the water and this metal is 53°C instead of 39°C, what quantity of this metal (initially at 125°C) was dropped into the insulated beaker?

(a)

Expert Solution

To determine

The amount of heat transferred from metal to water.

Answer to Problem 4SP

The amount of heat transferred from metal to water is 1932 cal.

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°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 8 4g for m, 1 cal/g⋅°C for c, 16°C for T1 and 39°C for T2 in equation (3)

Q=84 g×1 cal/g⋅°C(39°C−16°C)=1932 cal

Conclusion:

The amount of heat transferred from metal to water is 1932 cal.

(b)

Expert Solution

To determine

The specific heat capacity of metal.

Answer to Problem 4SP

The specific heat capacity of metal is 0.132 cal/g⋅°C.

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°C.

Write the expression for specific heat capacity of metal

c=Qm(T2−T1)

Substitute 170 g for m, 1932 cal for Q, 125°C for T1, and 39°C for T2 in the above equation

c=1932 cal170 g(39°C−125°C)=1932 cal170 g×(86°C)=0.132 cal/g⋅°C

The negative sign in the above expression is avoided since specific heat capacity is a positive quantity.

Conclusion:

The specific heat capacity of metal is 0.132 cal/g⋅°C.

(c)

Expert Solution

To determine

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

Answer to Problem 4SP

The quantity of heat was dropped from metal to insulating beaker is 327 g.

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°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 84 g for m, 1 cal/g⋅°C for c, 16°C for T1 and 53°C for T2 in equation (3)

Q=84 g×1 cal/g⋅°C(53°C−16°C) =3108 cal

Write the expression of mass in terms of specific heat capacity

m=Qc(T2−T1) (4)

Substitute 0.132 cal/g⋅°C for c, 3108 cal for Q, 125°C for T1 and 53°C for T2 in equation (4)

m=3108 cal0.132 cal/g⋅°C×(53°C−125°C)=327 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 327 g.

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Answer 3

Textbook Question

Chapter 10, Problem 4SP

A 170-g quantity of a certain metal, initially at 125°C, is dropped into an insulated beaker containing 84 g of water at 16°C. The final temperature of the metal and water in the beaker is measured as 39°C. Assume that the heat capacity of the beaker can be ignored.

a. How much heat has been transferred from the metal to the water?

b. Given the temperature change and mass of the metal, what is the specific heat capacity of the metal?

c. If the final temperature of the water and this metal is 53°C instead of 39°C, what quantity of this metal (initially at 125°C) was dropped into the insulated beaker?

(a)

Expert Solution

To determine

The amount of heat transferred from metal to water.

Answer to Problem 4SP

The amount of heat transferred from metal to water is 1932 cal.

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°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 8 4g for m, 1 cal/g⋅°C for c, 16°C for T1 and 39°C for T2 in equation (3)

Q=84 g×1 cal/g⋅°C(39°C−16°C)=1932 cal

Conclusion:

The amount of heat transferred from metal to water is 1932 cal.

(b)

Expert Solution

To determine

The specific heat capacity of metal.

Answer to Problem 4SP

The specific heat capacity of metal is 0.132 cal/g⋅°C.

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°C.

Write the expression for specific heat capacity of metal

c=Qm(T2−T1)

Substitute 170 g for m, 1932 cal for Q, 125°C for T1, and 39°C for T2 in the above equation

c=1932 cal170 g(39°C−125°C)=1932 cal170 g×(86°C)=0.132 cal/g⋅°C

The negative sign in the above expression is avoided since specific heat capacity is a positive quantity.

Conclusion:

The specific heat capacity of metal is 0.132 cal/g⋅°C.

(c)

Expert Solution

To determine

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

Answer to Problem 4SP

The quantity of heat was dropped from metal to insulating beaker is 327 g.

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°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 84 g for m, 1 cal/g⋅°C for c, 16°C for T1 and 53°C for T2 in equation (3)

Q=84 g×1 cal/g⋅°C(53°C−16°C) =3108 cal

Write the expression of mass in terms of specific heat capacity

m=Qc(T2−T1) (4)

Substitute 0.132 cal/g⋅°C for c, 3108 cal for Q, 125°C for T1 and 53°C for T2 in equation (4)

m=3108 cal0.132 cal/g⋅°C×(53°C−125°C)=327 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 327 g.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 10, Problem 4SP

A 170-g quantity of a certain metal, initially at 125°C, is dropped into an insulated beaker containing 84 g of water at 16°C. The final temperature of the metal and water in the beaker is measured as 39°C. Assume that the heat capacity of the beaker can be ignored.

a. How much heat has been transferred from the metal to the water?

b. Given the temperature change and mass of the metal, what is the specific heat capacity of the metal?

c. If the final temperature of the water and this metal is 53°C instead of 39°C, what quantity of this metal (initially at 125°C) was dropped into the insulated beaker?

(a)

Expert Solution

To determine

The amount of heat transferred from metal to water.

Answer to Problem 4SP

The amount of heat transferred from metal to water is 1932 cal.

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°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 8 4g for m, 1 cal/g⋅°C for c, 16°C for T1 and 39°C for T2 in equation (3)

Q=84 g×1 cal/g⋅°C(39°C−16°C)=1932 cal

Conclusion:

The amount of heat transferred from metal to water is 1932 cal.

(b)

Expert Solution

To determine

The specific heat capacity of metal.

Answer to Problem 4SP

The specific heat capacity of metal is 0.132 cal/g⋅°C.

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°C.

Write the expression for specific heat capacity of metal

c=Qm(T2−T1)

Substitute 170 g for m, 1932 cal for Q, 125°C for T1, and 39°C for T2 in the above equation

c=1932 cal170 g(39°C−125°C)=1932 cal170 g×(86°C)=0.132 cal/g⋅°C

The negative sign in the above expression is avoided since specific heat capacity is a positive quantity.

Conclusion:

The specific heat capacity of metal is 0.132 cal/g⋅°C.

(c)

Expert Solution

To determine

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

Answer to Problem 4SP

The quantity of heat was dropped from metal to insulating beaker is 327 g.

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°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 84 g for m, 1 cal/g⋅°C for c, 16°C for T1 and 53°C for T2 in equation (3)

Q=84 g×1 cal/g⋅°C(53°C−16°C) =3108 cal

Write the expression of mass in terms of specific heat capacity

m=Qc(T2−T1) (4)

Substitute 0.132 cal/g⋅°C for c, 3108 cal for Q, 125°C for T1 and 53°C for T2 in equation (4)

m=3108 cal0.132 cal/g⋅°C×(53°C−125°C)=327 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 327 g.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The amount of heat transferred from metal to water.

Answer to Problem 4SP

The amount of heat transferred from metal to water is 1932 cal.

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°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 8 4g for m, 1 cal/g⋅°C for c, 16°C for T1 and 39°C for T2 in equation (3)

Q=84 g×1 cal/g⋅°C(39°C−16°C)=1932 cal

Conclusion:

The amount of heat transferred from metal to water is 1932 cal.

(b)

Expert Solution

To determine

The specific heat capacity of metal.

Answer to Problem 4SP

The specific heat capacity of metal is 0.132 cal/g⋅°C.

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°C.

Write the expression for specific heat capacity of metal

c=Qm(T2−T1)

Substitute 170 g for m, 1932 cal for Q, 125°C for T1, and 39°C for T2 in the above equation

c=1932 cal170 g(39°C−125°C)=1932 cal170 g×(86°C)=0.132 cal/g⋅°C

The negative sign in the above expression is avoided since specific heat capacity is a positive quantity.

Conclusion:

The specific heat capacity of metal is 0.132 cal/g⋅°C.

(c)

Expert Solution

To determine

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

Answer to Problem 4SP

The quantity of heat was dropped from metal to insulating beaker is 327 g.

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°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 84 g for m, 1 cal/g⋅°C for c, 16°C for T1 and 53°C for T2 in equation (3)

Q=84 g×1 cal/g⋅°C(53°C−16°C) =3108 cal

Write the expression of mass in terms of specific heat capacity

m=Qc(T2−T1) (4)

Substitute 0.132 cal/g⋅°C for c, 3108 cal for Q, 125°C for T1 and 53°C for T2 in equation (4)

m=3108 cal0.132 cal/g⋅°C×(53°C−125°C)=327 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 327 g.

Answer 8

(a)

Expert Solution

To determine

The amount of heat transferred from metal to water.

Answer to Problem 4SP

The amount of heat transferred from metal to water is 1932 cal.

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°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 8 4g for m, 1 cal/g⋅°C for c, 16°C for T1 and 39°C for T2 in equation (3)

Q=84 g×1 cal/g⋅°C(39°C−16°C)=1932 cal

Conclusion:

The amount of heat transferred from metal to water is 1932 cal.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The amount of heat transferred from metal to water.

Answer 13

Answer to Problem 4SP

The amount of heat transferred from metal to water is 1932 cal.

Answer 14

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°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 8 4g for m, 1 cal/g⋅°C for c, 16°C for T1 and 39°C for T2 in equation (3)

Q=84 g×1 cal/g⋅°C(39°C−16°C)=1932 cal

Conclusion:

The amount of heat transferred from metal to water is 1932 cal.

Answer 15

(b)

Expert Solution

To determine

The specific heat capacity of metal.

Answer to Problem 4SP

The specific heat capacity of metal is 0.132 cal/g⋅°C.

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°C.

Write the expression for specific heat capacity of metal

c=Qm(T2−T1)

Substitute 170 g for m, 1932 cal for Q, 125°C for T1, and 39°C for T2 in the above equation

c=1932 cal170 g(39°C−125°C)=1932 cal170 g×(86°C)=0.132 cal/g⋅°C

The negative sign in the above expression is avoided since specific heat capacity is a positive quantity.

Conclusion:

The specific heat capacity of metal is 0.132 cal/g⋅°C.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The specific heat capacity of metal.

Answer 20

Answer to Problem 4SP

The specific heat capacity of metal is 0.132 cal/g⋅°C.

Answer 21

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°C.

Write the expression for specific heat capacity of metal

c=Qm(T2−T1)

Substitute 170 g for m, 1932 cal for Q, 125°C for T1, and 39°C for T2 in the above equation

c=1932 cal170 g(39°C−125°C)=1932 cal170 g×(86°C)=0.132 cal/g⋅°C

The negative sign in the above expression is avoided since specific heat capacity is a positive quantity.

Conclusion:

The specific heat capacity of metal is 0.132 cal/g⋅°C.

Answer 22

(c)

Expert Solution

To determine

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

Answer to Problem 4SP

The quantity of heat was dropped from metal to insulating beaker is 327 g.

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°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 84 g for m, 1 cal/g⋅°C for c, 16°C for T1 and 53°C for T2 in equation (3)

Q=84 g×1 cal/g⋅°C(53°C−16°C) =3108 cal

Write the expression of mass in terms of specific heat capacity

m=Qc(T2−T1) (4)

Substitute 0.132 cal/g⋅°C for c, 3108 cal for Q, 125°C for T1 and 53°C for T2 in equation (4)

m=3108 cal0.132 cal/g⋅°C×(53°C−125°C)=327 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 327 g.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

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

Answer 27

Answer to Problem 4SP

The quantity of heat was dropped from metal to insulating beaker is 327 g.

Answer 28

Explanation of Solution

Given info: A 170 g quantity of a certain metal, initially at 125°C , is dropped into an insulated beaker containing 84 g of water at 16°C . The final temperature of the metal and water in the beaker is measured as 39°C.