91Ó°ÊÓ

The power of the Carnot engine,P=500 W

The low temperature of the heat reservoir,${\mathrm{T}}_{\mathrm{I}}=60°\mathrm{C}=333\mathrm{K}$

The high temperature of the heat reservoir, ${\mathrm{T}}_{\mathrm{H}}=100°\mathrm{C}=373\mathrm{K}$,

By using the efficiency and temperature relation, we can calculate the total efficiency, and after that, by comparing the work per unit of heat, we can get the heat input from the work done, and we can find the rate of exhaust heat output.

Formulae:

The efficiency of the Carnot cycle,

$\mathrm{Î\mu }=\frac{\left({\mathrm{T}}_{\mathrm{H}}-{\mathrm{T}}_{\mathrm{L}}\right)}{{\mathrm{T}}_{\mathrm{H}}}=\frac{\left|\mathrm{W}\right|}{{\mathrm{Q}}_{\mathrm{H}}}$ (1)

The work is done per cycle using the first law of thermodynamics,

$\mathrm{W}={\mathrm{Q}}_{\mathrm{H}}-{\mathrm{Q}}_{\mathrm{L}}$ (2)

By substituting the given values in equation (1), we can get the efficiency of the engine as given:

$$\begin{gathered} \mathrm{Î\mu }=\frac{373\mathrm{K}-333\mathrm{K}}{373\mathrm{K}} \\ =\frac{40\mathrm{K}}{373\mathrm{K}} \\ =0.11 \end{gathered}$$

Now, using the same efficiency in the equation (1), we can get the energy transferred as heat is given as follows:

${\mathrm{Q}}_{\mathrm{H}}=\frac{1}{\mathrm{Î\mu }}×\mathrm{W}$

Now, differentiating the above equation with respect to time, we can get the rate of input heat as given:

$$\begin{gathered} \frac{{\mathrm{dQ}}_{\mathrm{H}}}{\mathrm{dt}}=\frac{1}{\mathrm{Î\mu }}×\frac{\mathrm{dW}}{\mathrm{dt}} \\ =\frac{1}{\mathrm{Î\mu }}×\mathrm{P} \\ =\frac{1}{0.11}×500 \\ =4.66×{10}^3\mathrm{J}/\mathrm{s} \end{gathered}$$

Hence, the value of the rate of input heat is$4.66×{10}^3\mathrm{J}/\mathrm{s}$

Now, substituting the value of equation (2) in equation (1), we get the value of efficiency as:

$$\begin{gathered} \mathrm{Î\mu }=\frac{\left|{\mathrm{Q}}_{\mathrm{H}}\right|-\left|{\mathrm{Q}}_{\mathrm{L}}\right|}{\left|{\mathrm{Q}}_{\mathrm{H}}\right|} \\ =1-\frac{\left|{\mathrm{Q}}_{\mathrm{H}}\right|}{\left|{\mathrm{Q}}_{\mathrm{H}}\right|} \end{gathered}$$

where ${\mathrm{Q}}_{\mathrm{L}}$is the exhaust heat

By putting the value of efficiency in the above equation, we get the relation of input and exhaust heat as follows:

$$\begin{gathered} 0.11=1-\frac{\left|{\mathrm{Q}}_{\mathrm{L}}\right|}{\left|{\mathrm{Q}}_{\mathrm{H}}\right|} \\ \frac{\left|{\mathrm{Q}}_{\mathrm{L}}\right|}{\left|{\mathrm{Q}}_{\mathrm{H}}\right|}=1-0.11 \\ {\mathrm{Q}}_{\mathrm{L}}=0.89{\mathrm{Q}}_{\mathrm{H}} \end{gathered}$$

Differentiating the above equation with respect to time, we can get the rate of exhaust heat as follows:

$$\begin{gathered} \frac{\mathrm{dQ}}{\mathrm{dt}}=0.89\frac{{\mathrm{dQ}}_{\mathrm{H}}}{\mathrm{dt}} \\ =0.89×\left(4.66×{10}^3\mathrm{J}/\mathrm{s}\right) \\ =4.16×{10}^3\mathrm{J}/\mathrm{s} \end{gathered}$$

Hence, the value of the exhaust heat is $4.16×{10}^3\mathrm{J}/\mathrm{s}$