Question

A Carnot engine operating between energy reservoirs at temperatures $500\mathrm{k}$and $300\mathrm{k}$produces a power output of $1000\mathrm{W}$. What are (a) the thermal efficiency of this engine, (b) the rate of heat input, in $\mathrm{W}$and (c) the rate of heat output, in $\mathrm{W}$?

Short answer

a.Thermal efficiency of engine is $40\%$.

b.The rate of heat input is$2500\mathrm{W}$.

c.The rate of heat output is$1500\mathrm{W}$.

Step-by-step solution

Calculation of thermal efficiency of carnot engine (part a)

a.

The thermal efficiency of a Carnot engine working between the temperatures${\mathrm{T}}_{\mathrm{h}}$of the heat bath and${\mathrm{T}}_{\mathrm{c}}$is given as,

$\mathrm{\eta }=1-\frac{{\mathrm{T}}_{\mathrm{c}}}{{\mathrm{T}}_{\mathrm{h}}}$

Efficiency,

$\mathrm{\eta }=1-\frac{300}{500}=40\%$

Calculation of rate of heat input (part b)

b.

The efficiency can be expressed in terms of the power as,

$\mathrm{\eta }=\frac{\mathrm{W}}{{\mathrm{Q}}_{\mathrm{h}}}=\frac{\mathrm{Pt}}{{\mathrm{P}}_{\mathrm{h}}\mathrm{t}}=\frac{\mathrm{P}}{{\mathrm{P}}_{\mathrm{h}}}$

where $\mathrm{P}$is the useful power, W is work output.

${\mathrm{P}}_h$is the heating power required.

${\mathrm{P}}_h=\frac{\mathrm{P}}{\mathrm{\eta }}$

${\mathrm{P}}_h=\frac{1000}{0.4}=2500\mathrm{W}$

Calculation of rate of heat output (part c)

(c)

Using solution b,

${\mathrm{Q}}_{\mathrm{h}}=\mathrm{W}+{\mathrm{Q}}_{\mathrm{c}}$

Q_cis heat dumped to cold reservoir.

This might be written in terms of power as,

${\mathrm{P}}_{\mathrm{h}}=\mathrm{P}+{\mathrm{P}}_{\mathrm{c}}$

P is power output.

${\mathrm{P}}_{\mathrm{c}}={\mathrm{P}}_{\mathrm{h}}-\mathrm{P}$

${\mathrm{P}}_{\mathrm{c}}=2500-1000=1500\mathrm{W}$