What is the final temperature ( 𝑇f ) of the two objects after thermal equilibrium is reached?

Question

A 981 g copper pan is removed from the stove at a temperature of 150 °C. It is then stacked on top of a 687 g aluminum sheet at room temperature, 25 °C. The specific heat capacity for copper is 0.387 Jg·K and that for aluminum is 0.904 Jg·K. Assume no heat is gained or lost to the surroundings.

T_f=_____

Answer 1

According to the law of conservation of energy

Heat lost by hot copper MUST equal heat gained by cooler aluminum

heat lost by copper = q = mC∆T

m = mass of Cu = 981 g; C = sp.heat of Cu = 0.387 J/gº; ∆T = change in temperature of Cu = 150º-Tf)

heat gained by aluminum = q = mC∆T

m = mass of Al = 687 g; C = sp.heat Al = 0.904 J/gº; ∆T = change in temperature of Al = Tf-25º)

Setting the heat lost equal to heat gained, and solving for Tf, we have...

(981 g)(0.387 J/gº)(150-Tf) = (687 g)(0.904 J/gº)(Tf-25)

56,947 - 380Tf = 621Tf - 15,526

1001Tf = 72,473

Tf = 72.4ºC

NOTES:

Many will use -q Cu = +q Al. I avoid using negative signs and simply make ∆T a positive number in both cases.

The fact that C is given in units of Kelvin (temperature) doesn't matter since ∆T is a CHANGE in temperature and will be the same regardless of which temperature scale one uses. No need to change from C to K in these problems.