Question

(a) A Carnot engine operates between a hot reservoir at 320 Kand a cold one at 260 K. If the engine absorbs 500 Jas heat per cycle at the hot reservoir, how much work per cycle does it deliver? (b) If the engine working in reverses functions as a refrigerator between the same two reservoirs, how much work per cycle must be supplied to remove 1000Jas heat from the cold reservoir?

Short answer

1. The work per cycle of the Carnot engine is 93.8 J .
2. The work per cycle supplied to remove 1000J as heat from the cold reservoir is 231J.

Step-by-step solution

The given data

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,${\mathrm{Q}}_{\mathrm{L}}=1000\mathrm{J}$ .

Understanding the concept of the Carnot cycle

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$\epsilon$for any engine,$\mathrm{\epsilon }=\frac{\left|\mathrm{W}\right|}{\left|{\mathrm{Q}}_{\mathrm{H}}\right|}$ (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, ${\mathrm{\epsilon }}_{\mathrm{C}}=\left(1-\frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}}\right)$ (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)

a) Calculation of the work per cycle of the Carnot engine

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

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

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

b) Calculation of the work supplied to remove 1000J as heat

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$231\mathrm{J}$