A 0.10 kg piece of copper at an initial temperature of 95?C is dropped
into 0.20 kg of water contained in a 0.28 kg aluminum calorimeter. The
water and calorimeter are initially at 15?C. What is the final
temperature of the system when it reaches equilibrium?

---

Request for Question Clarification by hedgie-ga on 22 Jul 2006 20:36 PDT

A 0.10 kg + 0.20 kg  = .30 kg

may not fit into 0.28 kg aluminum calorimeter

Anyway, what does 0.28 kg calorimeter mean?

 It has a mass of .28 kg (when empty? when filled with what?)

A would give D- to a teacher who dreamed up this homework.

Again a simple heat balance question

This is typical data from calorimeter experiments commonly performed in
introductory labs:

Heat Loss = Heat Gain

Specific heat Al = .22 cal/gm deg C
Specific heat Cu = .093 cal/gm deg C
Heat Loss = (95 - Tfinal) * .093 cal/gm * 100g

Heat Gain = (Tfinal - 15) [(200g * 1 cal/gm) + (280gm * .22 cal/gm)]

(I left out the units for temperature)
I obtained 19 deg C for the final temp but you better check the math
(Actually 19.05 but you only have 2 significant figures for the specific heats.