Question

A Carnot engine whose high-temperature reservoir is at 400 Khas an efficiency of 30.0% . By how much should the temperature of the low-temperature reservoir be changed to increase the efficiency to 40.0 %?

Short answer

The change in temperature of the low-temperature reservoir to increase the efficiency to 40.0% is -40 K

Step-by-step solution

The given data

The original efficiency of the engine,${\mathrm{\epsilon }}_{\mathrm{C}}=30\%\mathrm{or}0.30$

The high temperature of the reservoir,${\mathrm{T}}_{\mathrm{H}}=400\mathrm{K}$

The increased efficiency of the engine, $\epsilon {\text{'}}_C=40\%\mathrm{or}0.40$

Understanding the concept of the Carnot engine

A Carnot engine is an ideal engine, that follows the cycle of fig 20 - 9and the efficiency can be given by the equation 20 - 13. ${\mathbf{T}}_{\mathbf{H}}$and ${\mathbf{T}}_{\mathbf{L}}$are the temperatures of the high and low-temperature reservoirs.Using the efficiency equation, we can calculate the change in ${\mathbf{T}}_{\mathbf{L}}$.

Formula:

The efficiency of the Carnot engine,

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

Calculation of the change in low-temperature of the reservoir

The equation linearly depends on ${\mathrm{T}}_{\mathrm{L}}$, so we can take the derivative of efficiency from equation (1), and the given values with respect to ${\mathrm{T}}_{\mathrm{L}}$is given as:

$\frac{{\mathrm{d\epsilon }}_{\mathrm{C}}}{{\mathrm{dT}}_{\mathrm{L}}}=-\frac{1}{{\mathrm{T}}_{\mathrm{H}}}$

This equation can be written as:

$$\begin{gathered} \frac{∆{\mathrm{\epsilon }}_{\mathrm{c}}}{∆{\mathrm{T}}_{\mathrm{L}}}=-\frac{1}{{\mathrm{T}}_{\mathrm{H}}} \\ \frac{0.40-0.30}{∆{\mathrm{T}}_{\mathrm{L}}}=\frac{-1}{400\mathrm{K}} \\ \frac{0.10}{∆\mathrm{T}}=\frac{-1}{400\mathrm{K}} \\ ∆{\mathrm{T}}_{\mathrm{L}}=-0.10\times 400\mathrm{K} \\ ∆{\mathrm{T}}_{\mathrm{L}}=-40\mathrm{K} \end{gathered}$$

Hence, the value of the change in the low-temperature reservoir is -40 K