word problem

Question

A 102.5g sample of an uknown metal was heated in a water bath of 99.5 celsius. The metal was than added to 25.7g of water in an insulated container and the temperature of the water changed form an initial temperature of 20.0 celsius to a final temperature of 41.5 celsius.

 

A. How much heat did the water gain.

 

B. How much heat did the metal lose.

 

C. Calculate the specific heat of the metal.

Answer 1

The key facts you need to know are:

 

- Change in heat = change in temperature x mass x specific heat

- Specific heat of water is 1

 

So...

 

 

A. The water temperature rose 21.5 °C, the mass of the water was 25.7 grams.  Since there was no change of state, the water gained 21.5 * 25.7 * 1 = 552.55 calories.

 

B. Because heat is energyy and energy is conserved, the metal gained exactly as much heat as the water lost, 552.55 calories.

 

C.  The temperature of the metal dropped from 99.5 to 41.5 °C, so it lost 58 °C.  It's mass was 102.5 g.  Let c = the specific heat of the metal.  So 58 * 102.5 * c = 552.55.  Solve that for c and you get 0.0929 as the specific heat of the metal.