91Ó°ÊÓ

The temperature at the hot reservoir is ${\mathrm{T}}_{\mathrm{H}}=320\mathrm{K}$.

The temperature at the cold reservoir is ${\mathrm{T}}_{\mathrm{L}}=260\mathrm{K}$.

The heat is absorbed in the hot reservoir, localid="1661334216062" ${\mathrm{Q}}_{\mathrm{H}}=500\mathrm{J}$.

The heat removes from the cold reservoir,${\mathrm{Q}}_{\mathrm{L}}=1000\mathrm{J}$ .

By combining Equations 20-11 and 20-13, we can find the work per cycle of the Carnot engine. Also by combining Equations 20-14 and 20-16, we can find the work per cycle supplied to remove 1000J as heat from the cold reservoir.

Formulae:

From Equation 20-11, the efficiency$\mathrm{Î\mu }$for any engine,$\mathrm{Î\mu }=\frac{\left|\mathrm{W}\right|}{\left|{\mathrm{Q}}_{\mathrm{H}}\right|}$ (1)

where, W is work done, and ${\mathrm{Q}}_{\mathrm{H}}$is the heat absorbed at the hot reservoir.

From Equation 20-13, the efficiency of the Carnot engine, ${\mathrm{Î\mu }}_{\mathrm{C}}=\left(1-\frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}}\right)$ (2)

where ${\mathrm{T}}_{\mathrm{L}}$is the temperature cold reservoir and ${\mathrm{T}}_{\mathrm{H}}$is the temperature at the hot reservoir.

From Equation 20-14, the efficiency of performance for any refrigerator,$\mathrm{K}=\frac{\left|{\mathrm{Q}}_{\mathrm{L}}\right|}{\mathrm{W}}$ (3)

where${\mathrm{Q}}_{\mathrm{L}}$ is theheat removed from the cold reservoir.

From Equation 20-16, the efficiency of performance for Carnot refrigerator,

${\mathrm{K}}_{\mathrm{C}}=\left(\frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}-{\mathrm{T}}_{\mathrm{L}}}\right)$ (4)

Comparing equations (1) and (2) and substituting the given values, we can get the work done per cycle of the Carnot engine given as:

$$\begin{gathered} \left|\mathrm{W}\right|=\left|{\mathrm{Q}}_{\mathrm{H}}\right|\left(1-\frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}}\right) \\ =500\mathrm{J}×\left(1-\frac{260\mathrm{K}}{320\mathrm{K}}\right) \\ =93.8\mathrm{J} \end{gathered}$$

Hence, the work done by the cycle is$93.8\mathrm{J}$

By combiningequations (3) and (4) and using the given values, we get the work supplied as heat is given as:

$$\begin{gathered} \left|\mathrm{W}\right|=\frac{\left|{\mathrm{Q}}_{\mathrm{L}}\right|}{\left({\displaystyle \frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}-{\mathrm{T}}_{\mathrm{L}}}}\right)} \\ =\frac{1000\mathrm{J}}{\left({\displaystyle \frac{260\mathrm{K}}{320\mathrm{K}-260\mathrm{K}}}\right)} \\ =231\mathrm{J} \end{gathered}$$

Hence, the value of the work supplied is$231\mathrm{J}$