Question

The specific heat of apples is about $3.7kJ/(kg.K)$. Estimate the entropy decrease in a $2.0kg$ bag of apples when it is cooled from room temperature by a refrigerator down to $4^{0}C$.

• A. $-0.47\frac{kJ}{K}$
• B. $-0.39\frac{kJ}{K}$
• C. $-0.59\frac{kJ}{K}$
• D. $-0.139\frac{kJ}{K}$

Answer 1

The correct option is A $-0.47\frac{kJ}{K}$

Room temperature is about ${22}^{0}C$ or $295K$. The refrigerated temperature corresponds to $277K$.

Thus, $s_2-s_1=m{c}_{v}ln\left(\frac{T_2}{T_1}\right)$$=\left(2.0kg\right)\left(3.7\frac{kJ}{kg.K}\right)ln\left(\frac{277K}{295K}\right)=-0.47\frac{kJ}{K}$

The entropy change is negative since the apples are being cooled down.

Notice that the answer is substantially different (and wrong) if you fail to convert the temperature to an absolute scale.