Question

A sheet of gold weighing 13.6 g and at a temperature of 16.7°C is placed flat on a sheet of iron weighing 24.5 g and at a temperature of 46.4°C. What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings. (Hint: The heat gained by the gold must be equal to the heat lost by the iron.)

Answer 1

Answer:

$T_f=42.3^{\circ}C$

Explanation:

The heat released by the iron is absorbed by the gold. This quantity is calculated with the formula $Q=mc\Delta T$, which tells us the amount of heat (energy) Q needed to increase the temperature of a material of mass m, specific heat c by a temperature difference$\Delta T$.

Since the heat released (which we will take as negative Q) by the iron (Fe) is absorbed (positive Q then) by the gold (Au) we write $Q_{Au}=-Q_{Fe}$, which means $m_{Au}c_{Au}\Delta T_{Au}=-m_{Fe}c_{Fe}\Delta T_{Fe}$, which is to say $m_{Au}c_{Au}(T_{fAu}-T_{0Au})=-m_{Fe}c_{Fe}(T_{fFe}-T_{0Fe})$.

We know that thermal equilibrium has been reached, which means that at the end everything has the same temperature, so we have $T_{fAu}=T_{fFe}=T_f$, and we can substitute $m_{Au}c_{Au}(T_f-T_{0Au})=-m_{Fe}c_{Fe}(T_f-T_{0Fe})$ and do $m_{Au}c_{Au}T_f-m_{Au}c_{Au}T_{0Au}=-m_{Fe}c_{Fe}T_f+m_{Fe}c_{Fe}T_{0Fe}$, to get finally $m_{Fe}c_{Fe}T_f+m_{Au}c_{Au}T_f=m_{Fe}c_{Fe}T_{0Fe}+m_{Au}c_{Au}T_{0Au}$, which means $(m_{Fe}c_{Fe}+m_{Au}c_{Au})T_f=m_{Fe}c_{Fe}T_{0Fe}+m_{Au}c_{Au}T_{0Au}$, giving us the final formula $T_f=\frac{m_{Fe}c_{Fe}T_{0Fe}+m_{Au}c_{Au}T_{0Au}}{m_{Fe}c_{Fe}+m_{Au}c_{Au}}$.

We can get from tables that $c_{Fe}=0.450J/g^{\circ}C$ and $c_{Au}=0.129J/g^{\circ}C$, so we put everything in a calculator and we get $T_f=\frac{(24.5g)(0.450J/g^{\circ}C)(46.4^{\circ}C)+(13.6g)(0.129J/g^{\circ}C)(16.7^{\circ}C)}{(24.5g)(0.450J/g^{\circ}C)+(13.6g)(0.129J/g^{\circ}C)}=42.3^{\circ}C$