Question

entropy change during heat transfer
heat flows from a steel object of mass m = 4.5 kg whose temperature is
th = 425 k to an identical object at tc = 295 k. assuming that the
objects are thermally isolated from the environment, what is the net
entropy change of the universe after thermal equilibrium has been
reached?
rightarrow delta s(\text{h})=int_{t_{h}}^{t}\frac{dq}{t}=mcint_{t_{h}}^{t}\frac{dt}{t}=mcln\frac{t}{t_{h}}
=4.5\times900\timeslnleft|\frac{300}{425}
ight|lt0
j/k
rightarrow delta s(\text{cold})=int_{t_{c}}^{t}\frac{dq}{t}=mcint_{t_{c}}^{t}\frac{dt}{t}=mclnleft|\frac{t}{t_{c}}
ight|
=4.5\times900\timeslnleft|\frac{300}{295}
ight|=? j/k gt0
delta s(\text{system})=delta s(\text{hot})+delta s(\text{cold})=(-)+(+) gt0

Explanation:

Step1: Calculate entropy change of hot object

The formula for entropy change of an object during heat - transfer is $\Delta S = mc\ln\frac{T_2}{T_1}$. For the hot steel object with $m = 4.5$ kg, $c=900$ J/(kg·K), $T_1 = 425$ K and $T_2 = 300$ K. So, $\Delta S_{hot}=mc\ln\frac{T_2}{T_1}=4.5\times900\times\ln\frac{300}{425}$.
$\Delta S_{hot}=4050\times\ln\frac{300}{425}\approx4050\times(- 0.347)\approx - 1405.35$ J/K.

Step2: Calculate entropy change of cold object

For the cold steel object with $m = 4.5$ kg, $c = 900$ J/(kg·K), $T_1 = 295$ K and $T_2 = 300$ K. So, $\Delta S_{cold}=mc\ln\frac{T_2}{T_1}=4.5\times900\times\ln\frac{300}{295}$.
$\Delta S_{cold}=4050\times\ln\frac{300}{295}\approx4050\times0.0167\approx67.64$ J/K.

Step3: Calculate net entropy change of the universe

The net entropy change of the universe $\Delta S_{universe}=\Delta S_{hot}+\Delta S_{cold}$.
$\Delta S_{universe}=- 1405.35 + 67.64=-1337.71$ J/K. But we made a mistake above. Since the two objects are identical and thermally isolated, at thermal equilibrium, the final temperature $T_f=\frac{T_{h}+T_{c}}{2}=\frac{425 + 295}{2}=360$ K.
For the hot object: $\Delta S_{hot}=mc\ln\frac{T_f}{T_{h}}=4.5\times900\times\ln\frac{360}{425}=4050\times\ln\frac{360}{425}\approx4050\times(-0.167)\approx - 676.35$ J/K.
For the cold object: $\Delta S_{cold}=mc\ln\frac{T_f}{T_{c}}=4.5\times900\times\ln\frac{360}{295}=4050\times\ln\frac{360}{295}\approx4050\times0.195\approx789.75$ J/K.
$\Delta S_{universe}=\Delta S_{hot}+\Delta S_{cold}=-676.35 + 789.75 = 113.4$ J/K.

Answer:

$113.4$ J/K