The engine that powers a crane burns fuel at a flame temperature of 2000°C. It is cooled by 20°C air. The crane lifts a 2000 kg steel girder 30 m upward. How much heat energy is transferred to the engine by burning fuel if the engine is 40% as efficient as a Carnot engine?

Carnot engine temperatures

Temperature of heat source $=T_H=2000°C$

Temperature of heat sink $=T_C=20°C$

Mass $\left(m\right)=2000Kg$

Distance$\left(h\right)=30m$

Conversion of temperature into kelvin

We know that the temperature conversion is given by $T_{Kelvin}=T_{Celsius}+273K$

Therefore,

Conversion of Carnot engine temperature $T_C=20°C$

$$\begin{gathered} T_C=273+20=293K \\ {T}_H=2000°C \\ {T}_H=273+2000=2273K \end{gathered}$$

Calculation of efficiency of heat engine

Temperature of heat source $=T_H=500°C$

Temperature of heat sink $=T_C=5°C$

We know that the thermal efficiency of heat engine is the ratio of the work done by the engine to the heat drawn out of the hot reservoir.

Here, $T_H$is the temperature of hot reservoir that is source and $T_C$is the temperature of cold reservoir i.e. sink. The efficiency is given by

$\eta =1-\frac{T_C}{T_H}\left(I\right)$

Substituting the values of the temperature of the hot reservoir and cold reservoir in the above equation, the efficiency is given by $\eta =0.871$

It is given that the engine takes $40\%$as the Carnot engine

Therefore the efficiency of the heat engine is calculated as localid="1648491796049" ${\eta }_{heatengine}=0.4\times 0.871$

${\eta }_{heatengine}=0.348$

$\left(II\right)$

Calculation of the amount of heat energy absorbed.

Mass $\left(m\right)=2000Kg$

Distance $\left(h\right)=30m$

Now energy used to lift the mass $\left(m\right)$through a distance of $\left(h\right)$is given by $mgh$

Therefore, the energy used to lift the mass is calculated as $mgh=2000kg\times 9.8m/s^2\times 30m$

$mgh=588000J\left(III\right)$

The net work done is defined as the energy extracted from the hot reservoir times the efficiency of the Carnot engine.

Therefore, $W_{out}=\eta Q_H\left(IV\right)$

$Q_H=\frac{W_{out}}{\eta }$

Substituting the values of efficiency from equation $\left(II\right)$and the net work done per cycle $W_{out}$in the above equation, energy absorbed is calculated as

$$\begin{gathered} Q_H=\frac{588000}{0.348} \\ {Q}_H=1.68\times {10}^6=1.68KJ \end{gathered}$$

Conclusion :

The amount of heat energy transferred to the engine is$1.68KJ$