Question

Figure 20-32 represents a Carnot engine that works between temperatures ${\mathit{T}}_{\mathbf{1}}\mathbf{=}\mathbf{400}\mathit{K}$ and${\mathit{T}}_{\mathbf{2}}\mathbf{=}\mathbf{150}\mathit{K}$ and drives a Carnot refrigerator that works between temperatures ${\mathit{T}}_{\mathbf{3}}\mathbf{=}\mathbf{325}\mathit{K}\mathbf{}\mathbf{and}{\mathit{T}}_{\mathbf{4}}\mathbf{=}\mathbf{225}\mathit{K}$ . What is the ratio${\mathit{Q}}_{\mathbf{3}}\mathbf{/}{\mathit{Q}}_{\mathbf{1}}$ ?

Short answer

The value for the ratio ${\mathrm{Q}}_3/{\mathrm{Q}}_1\mathrm{is}2.03$.

Step-by-step solution

The given data

Temperatures at which the Carnot engine operates are $T_1=400K\mathrm{and}{T}_2=150K$.

Temperatures at which Carnot refrigerator operates are $T_3=325\mathrm{and}{T}_4=225K$.

Understanding the concept of Carnot engine and refrigerator

We are given enough information to calculate the coefficient of performance for the refrigerator. From the diagram, we can say that ${\mathit{Q}}_{\mathbf{3}}\mathbf{=}{\mathit{Q}}_{\mathbf{4}}\mathbf{+}\mathit{W}$. Now, work done by the engine is used to drive the refrigerator, so they will have the same W. We can substitute W in terms of ${\mathit{Q}}_{\mathbf{1}}\mathit{K}$. Now plugging the value for each, we can find the required ratio.

Formulae:

The coefficient of performance of Carnot engine or refrigerator,

$K_C=\frac{T_L}{T_H-T_L}$ (1)

The efficiency of the engine,

$\epsilon =\frac{T_L-T_H}{T_L}$ (2)

The work done per cycle of the engine,

W = QK (3)

Calculation of the ratio Q3/Q1 

From the diagram we can get the value heats as given:

$$\begin{gathered} Q_3=Q_4+W \\ {Q}_4=Q_3-W \end{gathered}$$

Dividing W at both the sides in the above equation, we get that

$$\begin{gathered} \frac{Q_4}{W}=\frac{(Q_3-W)}{W} \\ =\frac{Q_3}{W}-1..............\left(4\right) \end{gathered}$$

Using equations (1) and (3) in the diagram, we can the value as:

$\frac{Q_4}{W}=\frac{T_4}{T_3-T_4}.............\left(5\right)$

Comparing both equations (4) and (5), we can get that

$\frac{T_4}{T_3-T_4}=\frac{Q_3}{W}-1...............\left(6\right)$

Now, work done by the engine is used to drive the refrigerator.

So W will be the same for both and using equation (2) and equation (3), we get the work done by the engine is given as:

$$\begin{gathered} \frac{W}{Q_1}=\frac{T_1-T_2}{T_1} \\ W=Q_1\times \left(\frac{T_1-T_2}{T_1}\right)..........\left(7\right) \end{gathered}$$

So substituting equation (7) in equation (6), the equation becomes

$$\begin{gathered} \frac{T_4}{T_3-T_4}=\frac{Q_3}{Q_1\times \left({\displaystyle \frac{T_1-T_2}{T_1}}\right)} \\ \frac{T_4}{T_3-T_4}+1=\frac{Q_3}{Q_1}\times \left(\frac{T_1}{T_1-T_2}\right) \\ \frac{T_3}{T_3-T_4}=\frac{Q_3}{Q_1}\times \left(\frac{T_1}{T_1-T_2}\right) \\ \frac{Q_3}{Q_1}=\left(\frac{T_3}{T_3-T_4}\right)\times \left(\frac{T_1}{T_1-T_2}\right) \\ \frac{Q_3}{Q_1}=\left(\frac{325K}{325K-225K}\right)\times \left(\frac{400K-150K}{400K}\right) \\ \frac{Q_3}{Q_1}=2.03 \end{gathered}$$

Hence, the value of the ratio is 2.03