Question

With $\mathbf{2}\mathbf{.}\mathbf{56}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathbf{J}$ of heat transfer into this engine, a given cyclical heat engine can do only role="math" localid="1668687707360" $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{J}$of work (a) What is the engine’s efficiency (b) How much heat transfer to the environment takes place.

Short answer

Engine’s Efficiency is $0.0586$ and amount of heat transfer to the environment is $2.41\times {10}^6\mathrm{J}$.

Step-by-step solution

Introduction

Heat Engine is a system which converts heat into work by taking heat from hot body (source) to carry out some work and there by discharging some of it at a very low temperature to cold body (sink). First and second law of thermodynamics helps in the operation of heat engine.

Parameters given and relationships-

Work done by the heat engine $W=1.56\times {10}^5\mathrm{J}$

Heat transfer to the engine $Q_1=2.56\times {10}^6\mathrm{J}$

Engine efficiency (ɳ) =?

$h=\frac{W}{Q_1}$

Here

$$\begin{gathered} h-\mathrm{Efficiency}\mathrm{of}\mathrm{Heat}\mathrm{Engine} \\ W-\mathrm{Word}\mathrm{done}\mathrm{by}\mathrm{the}\mathrm{heat}\mathrm{engine} \\ {Q}_1-\mathrm{Heat}\mathrm{Transfer}\mathrm{to}\mathrm{the}\mathrm{Heat}\mathrm{Engine} \end{gathered}$$

Calculation of efficiency and heat transfer to the environment

For efficiency-

$$\begin{gathered} h=\frac{W}{Q_1} \\ =\frac{1.5\times {10}^5}{2.56\times {10}^6} \\ =.0586 \end{gathered}$$

Heat Transfer to the environment –

$$\begin{gathered} W=Q_1-Q_2 \\ {Q}_2=Q_1-W \\ {Q}_2=2.56\times {10}^6-1.50\times {10}^5 \\ {Q}_2=2.41\times {10}^6\mathrm{J} \end{gathered}$$

Therefore, efficiency of the engine is $0.0586$ and amount of heat transfer to the environment is $2.41\times {10}^6\mathrm{J}$.