A 55.0g sample of a metal at 100 degree C is dropped into 100.0g of water at 25.0 degrees the temperature of the water increases to 28.5 degrees C.

Question

Determine the specific heat of the metal when the specific heat of water is 4.184 J/g degrees C

Answer 1

The heat LOST by the hot metal MUST equal the heat GAINED by the cooler water.

q = mC∆T where q = heat; m = mass; C = specific heat; ∆T = change in temperature

heat lost by metal = q = mC∆T = (55.0g)(C)(100 - 28.5º) = (55g)(C)(71.5º) = 3932.5 C

heat gained by water = q = mC∆T = (100.0g)(4.184 J/gº)(28.5 - 25º) = 418.4(3.5) = 1464.4

Set these two equal to each other and solve for C:

3932.5 C = 1464.4

C = 0.372 J/gº