Assume a piece of hot metal is added to a calorimeter containing 50mL of water at 22C. The amount of heat released by the metal was determined to be 1000J, and the equilibrium temperature in the calorimeter was 26C. Remembering that the specific heat of water is 4.18 J/gC, how much heat in joules does the calorimeter gain?

Question

Assume a piece of hot metal is added to a calorimeter containing 50mL of water at 22C. The amount of heat released by the metal was determined to be 1000J, and the equilibrium temperature in the calorimeter was 26C. Remembering that the specific heat of water is 4.18 J/gC, how much heat in joules does the calorimeter gain?
The answer is 164J. How do you get this? I tried Mass x specific heat x delta T and got 836 J but this is wrong.

DrBob222 · Answer

The 836 is ok; you just didn't go far enough.
q = heat water absorbed + heat cal absorbed.
1000 = [mass H2O x sp.h. H2O x (Tfinal-Tinitial) + Qcal
1000 = (50 x 4.18 x 4) + Qcal
1000 = 836 + Qcal
Qcal = 1000-836 = 164

Lance · Answer

What is Qcal?

Lance · Answer

Thank you. This helps!

Damon · Answer

Qcal is the heat absorbed by the calorimeter itself.