A Carnot engine whose sink is at 300 K has an efficiency of $40 \%. $ By how much should the temperature of source be increased so as to increase its efficiency by $50 \%$ of original efficiency :-

Question

A real gas within a closed chamber at $ 27^\circ \text{C} $ undergoes the cyclic process as shown in the figure. The gas obeys the equation $ PV^3 = RT $ for the path A to B. The net work done in the complete cycle is (assuming $ R = 8 \, \text{J/molK} $):

Answer 1

Let's solve this problem step-by-step using the concepts of thermodynamics related to the Carnot engine.

The efficiency ($ \eta $) of a Carnot engine is given by the formula:

$\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}}$

where $ T_{\text{sink}} $ is the temperature of the sink and $ T_{\text{source}} $ is the temperature of the source. All temperatures need to be in Kelvin.

Given:

Sink temperature ($ T_{\text{sink}} $) = 300 K
Initial efficiency ($ \eta_{\text{initial}} $) = 40% = 0.4

We need to find the original source temperature ($ T_{\text{source, initial}} $) using the formula:

$0.4 = 1 - \frac{300}{T_{\text{source, initial}}}$

Rearranging gives:

$ \frac{300}{T_{\text{source, initial}}} = 0.6 $

$T_{\text{source, initial}} = \frac{300}{0.6} = 500 \, \text{K}$

Next, we need to increase the efficiency by 50% of the original efficiency. This means the new efficiency ($ \eta_{\text{new}} $) will be:

$ \eta_{\text{new}} = \eta_{\text{initial}} + 0.5 \times \eta_{\text{initial}} = 0.4 + 0.2 = 0.6 $

We now set up the equation for the new efficiency:

$0.6 = 1 - \frac{300}{T_{\text{source, new}}}$

Solving for the new source temperature ($ T_{\text{source, new}} $) gives:

$ \frac{300}{T_{\text{source, new}}} = 0.4 $

$T_{\text{source, new}} = \frac{300}{0.4} = 750 \, \text{K}$

Thus, the temperature of the source needs to be increased by:

$ \Delta T = T_{\text{source, new}} - T_{\text{source, initial}} = 750 - 500 = 250 \, \text{K}$

Therefore, the correct answer is 250 K.

Answer 2