Question

(a) During each cycle, a Carnot engine absorbs 750 Jas heat from a high-temperature reservoir at 360 K , with the low-temperature reservoir at 280 K . How much work is done per cycle? (b) The engine is then made to work in reverse to function as a Carnot refrigerator between those same two reservoirs. During each cycle, how much work is required to remove 1200Jas heat from the low-temperature reservoir?

Short answer

1. Work done per cycle of Carnot engine is 166.5 J
2. Work required to remove 1200 J as heat from the low-temperature reservoir is 342.8 J

Step-by-step solution

 Step 1: The given data

The high-temperature reservoir is at,${\mathrm{T}}_{\mathrm{H}}=360\mathrm{K}$

The low-temperature reservoir is at,${\mathrm{T}}_{\mathrm{L}}=280\mathrm{K}$

Heat absorbed from the high-temperature reservoir is${\mathrm{Q}}_{\mathrm{H}}=750\mathrm{J}$

Heat absorbed from the low-temperature reservoir is ${\mathrm{Q}}_{\mathrm{L}}=1200\mathrm{J}$

Understanding the concept of the Carnot engine

Using the equations for the efficiency of the Carnot engine, we can find the work done per cycle. We can find the work required to remove heat using the formulae for the coefficient of performance of the refrigerator.

Formulae:

The efficiency of the engine,

$\mathrm{\epsilon }=\frac{\left|\mathrm{W}\right|}{\left|{\mathrm{Q}}_{\mathrm{H}}\right|}$ (1)

The efficiency of the Carnot cycle,

$\mathrm{\epsilon }=1-\frac{\left|{\mathrm{Q}}_{\mathrm{L}}\right|}{\left|{\mathrm{Q}}_{\mathrm{H}}\right|}=1-\frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}}$ (2)

Coefficient of performance of refrigerator,

$\mathrm{K}=\frac{\left|{\mathrm{Q}}_{\mathrm{L}}\right|}{\left|\mathrm{W}\right|}$ (3)

Coefficient of performance for Carnot cycle,

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

(a) Calculation of work done per cycle of Carnot engine

From the equation (2) of efficiency of the Carnot cycle, we can get that

$$\begin{gathered} {\mathrm{\epsilon }}_{\mathrm{c}}=1-\frac{280\mathrm{K}}{360\mathrm{K}} \\ =0.222 \end{gathered}$$

Substituting the given values in equation (1), we can get the work done per cycle of the engine is given as:

$$\begin{gathered} 0.222=\frac{\left|\mathrm{W}\right|}{\left|750\right|} \\ \left|\mathrm{W}\right|=166.5\mathrm{J} \end{gathered}$$

Hence, the work done per cycle of the engine is 166.5 J

(b) Calculation of required work to remove as heat from the low-temperature reservoir

The coefficient of performance ${\mathrm{K}}_{\mathrm{c}}$for the Carnot cycle using equation (4) is given by:

$$\begin{gathered} {\mathrm{K}}_{\mathrm{c}}=\frac{280\mathrm{K}}{360\mathrm{K}-280\mathrm{K}} \\ =3.5 \end{gathered}$$

Since the Carnot engine is made to work as a refrigerator

So, the coefficient of performance K of the refrigerator using equation (3) is given as:

$$\begin{gathered} 3.5=\frac{\left|1200\right|}{\left|\mathrm{W}\right|} \\ \mathrm{W}=342.8\mathrm{J} \end{gathered}$$

Hence, the work done per cycle of the refrigerator is 342.8 J