Question

How much work must be done by a Carnot refrigerator to transfer 1.0 J as heat (a) from a reservoir at $\mathbf{7}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{C}$to one at $\mathbf{27}\mathbf{°}\mathit{C}$, (b) from a reservoir at $\mathbf{-}\mathbf{73}\mathbf{°}\mathit{C}$to one at $\mathbf{27}\mathbf{°}\mathit{C}$, (c) from a reservoir at $\mathbf{-}\mathbf{173}\mathbf{°}\mathit{C}$to one at $\mathbf{27}\mathbf{°}\mathit{C}$, and (d) from a reservoir at $\mathbf{-}\mathbf{233}\mathbf{°}\mathbf{C}$to one at $\mathbf{27}\mathbf{°}\mathit{C}$?

Short answer

1. The work done by the Carnot refrigerator from a reservoir at $7°C$to one at $27°C$is 0.071J
2. The work done by the Carnot refrigerator from a reservoir at$-73°C$to one at $27°C$is 0.50J.
3. The work done by the Carnot refrigerator from a reservoir at to one at $-173°C$to one at $27°C$is 2.0J.
4. The work done by the Carnot refrigerator from a reservoir at to one at$-233°C$to one at $27°C$is 5.0J.

Step-by-step solution

The given data

Energy transferred by the Carnot refrigerator as heat,$Q_L=1.0J$

Understanding the concept of the Carnot refrigerator

A Carnot refrigerator operates as a reversible Carnot cycle by rejecting the heat of the freezer to the environment. Thus, using the work done and its performance relation, we can get the required work values in each given case.

Formulae:

The coefficient of performance of a Carnot refrigerator,$K_C=\frac{T_L}{T_H-T_L}$ (1)

The work done per cycle of a Carnot refrigerator, $\left|W\right|=\frac{\left|Q_L\right|}{K_c}$ (2)

a) Calculation of the work done from 7oC to 27oC

Here, the low temperature of the reservoir is at $T_L=7°C=280K$and the high temperature is at $T_H=27°C=300K$

Now, the work done by the Carnot refrigerator from a reservoir at $7^{°}C\mathrm{to}\mathrm{one}\mathrm{at}27°C$is given by substituting equation (1) in equation (ii) with the given data as follows:

$$\begin{gathered} \left|\mathrm{W}\right|=\left|{\mathrm{Q}}_{\mathrm{L}}\right|\left(\frac{{\mathrm{T}}_{\mathrm{H}}-{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{L}}}\right) \\ =(1.0\mathrm{J})\left(\frac{300\mathrm{K}-280\mathrm{K}}{280\mathrm{K}}\right) \\ =0.071\mathrm{J} \end{gathered}$$

Hence, the value of the work done is 0.071J.

b) Calculation of the work done from -73oC to 27oC

Here, the low temperature of the reservoir is at$T_L=-{73}^{°}C=200K$and the high temperature is at$T_H={27}^{°}C=300K$

Now, the work done by the Carnot refrigerator from a reservoir at$-73°C$to one at$27°C$is given by substituting equation (1) in equation (2) with the given data as follows:

$$\begin{gathered} \left|\mathrm{W}\right|=\left|{\mathrm{Q}}_{\mathrm{L}}\right|\left(\frac{{\mathrm{T}}_{\mathrm{H}}-{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{L}}}\right) \\ =(1.0\mathrm{J})\left(\frac{300\mathrm{K}-200\mathrm{K}}{200\mathrm{K}}\right) \\ =0.50\mathrm{J} \end{gathered}$$

Hence, the value of the work done is 0.50 J.

c) Calculation of the work done from -173oC to 27oC

Here, the low temperature of the reservoir is at$T_L=-{173}^{°}C=100K$and the high temperature is at$T_H={27}^{°}C=300K$

Now, the work done by the Carnot refrigerator from a reservoir at$173°C$to one at$27°C$is given by substituting equation (1) in equation (2) with the given data as follows:

$$\begin{gathered} \left|\mathrm{W}\right|=\left|{\mathrm{Q}}_{\mathrm{L}}\right|\left(\frac{{\mathrm{T}}_{\mathrm{H}}-{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{L}}}\right) \\ =(1.0\mathrm{J})\left(\frac{300\mathrm{K}-100\mathrm{K}}{100\mathrm{K}}\right) \\ =2.0\mathrm{J} \end{gathered}$$

Hence, the value of the work done is 2.0 J.

d) Calculation of the work done from -223oC to 27oC

Here, the low temperature of the reservoir is at${\mathrm{T}}_{\mathrm{L}}=-223°\mathrm{C}=50\mathrm{K}$and the high temperature is at$T_H=27°C=300K$

Now, the work done by the Carnot refrigerator from a reservoir at$-233°C$to one at$27°C$is given by substituting equation (1) in equation (2) with the given data as follows:

$$\begin{gathered} \left|\mathrm{W}\right|=\left|{\mathrm{Q}}_{\mathrm{L}}\right|\left(\frac{{\mathrm{T}}_{\mathrm{H}}-{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{L}}}\right) \\ =(1.0\mathrm{J})\left(\frac{300\mathrm{K}-50\mathrm{K}}{50\mathrm{K}}\right) \\ =5.0\mathrm{J} \end{gathered}$$

Hence, the value of the work done is 5.0 J.