Chemistry......

Question

A 157.6 g piece of metal is placed into a beaker of boiling water. You record the temperature of boiling water to be 97.1 . The metal is then placed into a coffee cup calorimeter that contains contains 58.5 g of water. The initial temperature of the water in the coffee cup calorimeter is 12.4 . You measure the final temperature of the water in the coffee cup calorimeter to be 43.7. Determine the change in temperature of the water in the coffee cup calorimeter. Assume the specific heat capacity of water to be 4.18 J/g. 

Answer 1

The change in temperature in the coffee cup calorimeter is going to be 43.7º - 12.4º = 31.3º. Are you sure that's the only question you have for this problem? Seems rather simple, unless I'm missing something. Did you maybe need to find the specific heat of the metal? If so, here is how you find that.....

heat gained by water = q = mC∆T

q = (58.5 g)(4.18 J/gº)(31.3º) = 7654 J

heat lost by metal = q = mC∆T

q = (157.6 g)(C)(97.1º - 43.7º) and this must be equal to the heat gained by the water (conservation of energy)

Therefore...

7654 J = (157.6 g)(C)(53.4º) = 8416 gº(C)

C = specific heat of metal = 0.909 J/gº