chemistry calorimetry

Question

In the laboratory a student uses a "coffee cup" calorimeter to determine the specific heat of a metal.  

She heats 19.0 grams of copper to 99.29°C and then drops it into a cup containing 77.5 grams of water at 21.51°C. She measures the final temperature to be 23.16°C.  

Assuming that all of the heat is transferred to the water, she calculates the specific heat of copper to be  _____ J/g°C.

Answer 1

heat lost by copper = heat gained by water in the calorimeter

heat lost by copper = q = mC∆T = mass Cu x specific heat Cu x change in temperature of Cu

q = (19.0 g)(C)(23.16º - 99.29º) = -1444gºC

heat gained by water = q = mC∆T = mass H2O x specific heat H2O x change in temperature of H2O

q = (77.5 g)(4.184 J/gº)(23.16º -21.51º) = 535 J

1444gºC = 535 J

C = 535 J / 1446gº

C = 0.370 J/gº

Done another way, which may make more sense...

heat gained by water must equal heat lost by Cu

q = mC∆T

For water:

q = (77.5 g)(4.184 J/gº)(1.65º) = 535 J of heat gained by water

For Cu:

q = (19.0 g)(C)(76.13º) = heat lost by Cu

Setting them equal and solving for C of copper, we have ...

535 J = (19.0 g)(C)(76.13º) = 1446 gº (C)

C = 535 J / 1446 gº

C = 0.370 J/gº