Answer to Question #160502 in Molecular Physics | Thermodynamics for Reinon Arvi Layrit

Question

Jill added 215J of heat energy to 14.5 kg of copper and 1.50 kg of water which are in

thermal contact with each other, both initially at 32.0^oC. What will be their final temperature?

(C_copper 0.385 J/g^oC, C_water 4.186 J/g^oC).

Answer 1

$Q=c_{copper}\times{m_{copper}}\times\Delta{T}+c_{water}\times{m_{water}}\times\Delta{T}$

Let $x$ be the final temperature. Then the equation for the heat can be represented in the following way (the mass should be in grams):

$0.385\times14500\times(x-32)+4.186\times1500\times(x-32)=215$

Solving for x,

$x=32.018...\approx32.0$ (to three significant figures).

So the final temperature will be just slightly higher than the initial temperature, and since all the values are given to 3 significant digits, the final temperature should also be rounded to 3 significant figures. The reason of this result is that the heat of 215 Joules is negligible for such large amounts of substances (14.5 kg of copper and 1.5 kg of water), so the temperature of the system changes very insignificantly.

Answer: 32.0^oC