91Ó°ÊÓ

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

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)

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.

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.

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.

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.