Physics Problems & Solutions: Thermal Physics

A body of mass m with spesific heat C at temperature 500 K is brought into contact with an identical body at temperature 100 K, and the two are isolated from their surroundings. The change in entropy of the system is equal to

A. ⁴⁄₃ mC
B. mC ln ⁹⁄₅
C. mC ln 3
D. –mC ln ⁵⁄₃
E. 0

(GR9677 #74)

Solution:

Entropy:
$dS=\frac{dQ}{T}= \frac{mcdT}{T}\rightarrow S=mc\int_{T_i}^{T_f}\frac{dT}{T}$

$S_{\text{net}}=m_1c_1\int_{T_1}^{T_f}\frac{dT}{T}+m_2c_2\int_{T_2}^{T_f}\frac{dT}{T}$

m₁= m₂= m
c₁= c₂= C
T₁= 500
T₂= 100
T_f= (500 + 100)/2 = 300

$\begin{aligned} \\S_{\text{net}}&=mC\left (\int_{500}^{300}\frac{dT}{T}+\int_{100}^{300}\frac{dT}{T} \right )\\&=mC\left (\ln \frac{300}{500}+\ln \frac{300}{100} \right )\\&=mC \ln \left (\frac{3}{5}\cdot\frac{3}{1} \right ) \\&= mC\ln\frac{9}{5} \end{aligned}$

Answer: B

Physics Problems & Solutions

Thermal Physics - Entropy

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