Solved Problem on Thermodynamics

A heat engine works between a high-temperature reservoir at 100 °C and a lower-temperature reservoir at 50 °C, calculate:
a) The thermal efficiency of this machine;
b) The work done by the engine when receiving 10000 kcal of the high-temperature reservoir. Assume 1 cal = 4.2 J.

Problem data:

Temperature of the hot reservoir: t_h = 100 °C;
Temperature of the cold reservoir: t_c = 50 °C.

Problem diagram:

Figure 1

Solution

First, we must convert temperatures given in degrees Celsius (°C) to kelvins (K) and energy (heat) given in item (b) in kilocalories (kcal) to joules (J), used in the International System of Units (S.I.)

\[ \begin{gather} T_{h}=t_{q}+273=100+273=373\;\text{K}\\[10pt] T_{c}=t_{f}+273=50+273=323\;\text{K}\\[10pt] Q=10000\;\text{kcal}=10000\times 10^{3}\;\cancel{\text{cal}}\times \frac{4.2\;\text{J}}{1\;\cancel{\text{cal}}}=1\times 10^{7}\times 4.2\;\text{J}=4.2\times 10^{7}\;\text{J} \end{gather} \]

a) The thermal efficiency is given b

\[ \bbox[#99CCFF,10px] {\eta =\frac{T_{h}-T_{c}}{T_{h}}} \]

\[ \begin{gather} \eta =\frac{373-323}{373}\\ \eta =\frac{50}{373}\\ \eta\simeq 0.13=\frac{13}{100} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {\eta =13\;\text{%}} \]

b) Using the expression of the thermal efficiency as a function of work done W and the energy received Q

\[ \bbox[#99CCFF,10px] {\eta =\frac{W}{Q}} \]

\[ \begin{gather} W=\eta Q\\ W=0.13\times 4.2\times 10^{7}\\ W=5460000 \end{gather} \]

\[ \bbox[#FFCCCC,10px] {W=5.46\times 10^{6}\;\text{J}} \]

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