First Law of Thermodynamics
7.0 Heat engine
7.0 Heat engine
A heat engine is a device that converts heat energy into mechanical energy. Any heat engine works on a thermodynamic system, let's take a gas in a cylinder. The system takes ${Q_1}$ amount of energy from the source at ${T_1}$ temperature, converted into $W$ amount of work and the rest goes ${Q_2}$ goes to the sink at ${T_2}$. From the conservation of energy,$${Q_1} = W + {Q_2}$$
Efficiency of a heat engine $\left( \eta \right)$:
$$\begin{equation} \begin{aligned} \eta = \frac{Energy\ we\ get}{Energy\ we\ pay\ for} = \frac{Work\ output}{Heat\ input} \\ \Rightarrow \eta = \frac{W}{{{Q_1}}} = \frac{{{Q_1} - {Q_2}}}{{{Q_1}}} = 1 - \frac{{{Q_2}}}{{{Q_1}}} \\\end{aligned} \end{equation} $$
In general, We acn say
$$\eta = \frac{{\sum W }}{{\sum {{Q_{ + ve}}} }} = \frac{{\sum {{Q_{ + ve}}} - \left| {\sum {{Q_{ - ve}}} } \right|}}{{\sum {{Q_{ + ve}}} }} = 1 - \left| {\frac{{\sum {{Q_{ - ve}}} }}{{\sum {{Q_{ + ve}}} }}} \right|$$
Question 17. A heat engine operates between a cold reservoir at temperature ${T_2} = 300\;K$ and a hot reservoir at temperature ${T_1}$. It takes $200 J$ of heat from the hot reservoir and delivers $120J$ of heat to the cold reservoir in a cycle. What could be the minimum temperature of hot reservoir ?
Solution: The work done by the engine in a cycle is $$W = 200J - 120J = 80J$$The efficiency of the engine is $$\eta = \frac{W}{Q} = \frac{{80\,J}}{{200\,J}} = 0.40$$From carnot's theorem, no engine can have an efficiency greater than that of carnot engine.
Thus,$$\begin{equation} \begin{aligned} 0.40 \leqslant 1 - \frac{{{T_2}}}{{{T_1}}} = 1 - \frac{{300K}}{{{T_1}}} \\ or,\quad \quad \frac{{300K}}{{{T_1}}} \leqslant 1 - 0.40 = 0.60 \\ or,\quad \quad {T_1} \geqslant \frac{{300K}}{{0.60}} \\ or,\quad \quad {T_1} \geqslant 500K \\\end{aligned} \end{equation} $$The minimum temperature of hot reservoir has to be 500K.
