91Ó°ÊÓ

1. The work rate of the Carnot engine, P = 500 W
2. The high temperature of the reservoir,$T_H=100°\mathrm{C}=373\mathrm{K}$
3. The low temperature of the reservoir, $T_L=60.0°\mathrm{C}=333\mathrm{K}$

We can find the efficiency of the Carnot engine using the corresponding formula. Dividing power by efficiency will give us the rate at which energy is taken in by the engine as heat. From power and absorbed energy, we can easily find the rate at which energy is exhausted by the engine as heat.

Formulae:

The efficiency of a Carnot engine,

$\mathrm{Î\mu }=\frac{{\mathrm{T}}_{\mathrm{H}}-{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}}$ (1)

The net work done by a Carnot engine,

$W=\mathrm{Î\mu }{Q}_H$ (2)

Using equation (1), the efficiency of the Carnot engine can be given as:

$$\begin{gathered} \mathrm{Î\mu }=\frac{373\mathrm{K}-333\mathrm{K}}{373\mathrm{K}} \\ =0.107 \end{gathered}$$

Work done per cycle per second is, 500 W

Energy added in the form of heat per second can be given using equation (2) as follows:

$$\begin{gathered} \frac{Q}{t}=\frac{\mathrm{P}}{\mathrm{Î\mu }} \\ =\frac{500\mathrm{W}}{0.107} \\ =4672\mathrm{W} \end{gathered}$$

Hence, the required value of the rate of energy to be taken in as heat is 4672W

The rate at which energy is exhausted by the engine as heat can be given as:

$$\begin{gathered} Q_{lost}=4672\mathrm{W}-500\mathrm{W} \\ =4172\mathrm{W} \end{gathered}$$

Hence, the value of the rate of energy be exhausted as heat to the engine is 4172 W