A 100g block of metal with a specific heat of 0.2 cal/gC degree that is at 100 d

Question

A 100g block of metal with a specific heat of 0.2 cal/gC degree that is at 100 degree C is dropped into a large Styrofoam cup that contains 100g of water that is at 20 degree C.

A) Compare the heat lost/gained by the water and the metal block. A numerical value is not needed but explain your response.

B) Compare the temperature change of the water and the metal block. Which will change the temp the most? Numerical values are not needed at this point, but explain answer.

C) What is the final temperature of the mixture? Do a calculation and start with energy conservation

Explanation / Answer

Part A)

The heat gained by the water will be exactly the amount of heat lost by the metal block. All energy must be conserved.

Part B)
The water temp change will be much smaller than the block temp change since the specific heat of water is much higher. As a matter of fact, the specific heat of water is 5 times greater than that of the block in this question, so the change in temp of the block will be 5 times the change in temp of the water given that they have the same mass. Proof below.

Part C)

Apply Q = mc(delta T)

mc(delta T) = mc(delta T)

(.1)(.2)(100 - Tf) = (.1)(1)(Tf - 20)

2 - .02Tf = .1Tf - 2

4 = .12Tf

Tf = 33.3

(As a proof to part B, notice that the block went from 100 - 33.3 losing 66.7 degrees. The water went from 20 to 33.3 gaining 13.3 degrees. 13.3(5) = 66.5. That is 5 times less than the change in temp of the block)

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