Question

A number of Carnot engines are operated at identical cold reservoir temperatures (T_L). However, their hot reservoir temperatures are kept different. A graph of the efficiency of the engines versus hot reservoir temperature (T_H) is plotted. The correct graphical representation is

• A.
• B.
• C.
• D.

Answer 1

The Correct Option is B

Step 1: Recall the formula for the efficiency of a Carnot engine.
The efficiency ($\eta$) of a Carnot engine operating between a hot reservoir at temperature $T_H$ and a cold reservoir at temperature $T_L$ is given by:
$\eta = 1 - \frac{T_L}{T_H}$
where $T_H$ and $T_L$ are absolute temperatures (in Kelvin).

Step 2: Understand the given conditions.
In this problem, the cold reservoir temperature $T_L$ is kept constant, while the hot reservoir temperature $T_H$ is varied. We need to determine the graphical representation of the efficiency ($\eta$) as a function of the hot reservoir temperature ($T_H$).

Step 3: Analyze the behavior of efficiency as $T_H$ changes.
- When $T_H = T_L$:
$\eta = 1 - \frac{T_L}{T_L} = 1 - 1 = 0$
So, when the hot reservoir temperature is equal to the cold reservoir temperature, the efficiency is zero.
- As $T_H$ increases ($T_H > T_L$):
As $T_H$ increases, the ratio $\frac{T_L}{T_H}$ decreases because $T_L$ is constant and positive.
Since $\eta = 1 - \frac{T_L}{T_H}$, as $\frac{T_L}{T_H}$ decreases, the efficiency $\eta$ increases.
- As $T_H \rightarrow \infty$:
As $T_H$ approaches infinity, the term $\frac{T_L}{T_H}$ approaches 0.
Thus, $\eta \rightarrow 1 - 0 = 1$
The efficiency approaches 1 (or 100%) as the hot reservoir temperature becomes infinitely large, but it can never reach 1 for finite $T_H$.

Step 4: Determine the shape of the graph.
To see how the efficiency increases with $T_H$, let's consider the rate of change of efficiency with respect to $T_H$.
$\frac{d\eta}{dT_H} = \frac{d}{dT_H} \left( 1 - \frac{T_L}{T_H} \right) = -T_L \frac{d}{dT_H} (T_H^{-1}) = -T_L (-1) T_H^{-2} = \frac{T_L}{T_H^2}$
Since $T_L > 0$ and $T_H^2 > 0$, $\frac{d\eta}{dT_H} > 0$. 
This confirms that efficiency increases as $T_H$ increases.
The second derivative is:
$\frac{d^2\eta}{dT_H^2} = \frac{d}{dT_H} \left( \frac{T_L}{T_H^2} \right) = T_L \frac{d}{dT_H} (T_H^{-2}) = T_L (-2) T_H^{-3} = -\frac{2T_L}{T_H^3}$
Since $T_L > 0$ and $T_H^3 > 0$, $\frac{d^2\eta}{dT_H^2} < 0$. 
This means the rate of increase of efficiency decreases as $T_H$ increases. 
The curve is concave down.

Step 5: Match the analysis with the given graphs.
- Option (A) and (D) are incorrect because they show efficiency decreasing as $T_H$ increases beyond $T_L$.
- Option (C) is incorrect because it shows a linear increase in efficiency, whereas the relationship is not linear and should be concave down.
- Option (B) starts from zero efficiency at $T_H = T_L$, increases as $T_H$ increases, and the rate of increase decreases as $T_H$ increases (concave down shape), approaching a maximum value. This matches our analysis.

Final Answer: 
The correct graphical representation of the relationship between efficiency and $T_H$ is Option (B)

Answer 2

The efficiency of a Carnot engine is given by:

$\eta = 1 - \frac{T_L}{T_H}$

where:

• $T_L$ is the absolute temperature of the cold reservoir, and
• $T_H$ is the absolute temperature of the hot reservoir.

When $T_H = T_L$, the efficiency is $\eta = 1 - \frac{T_L}{T_L} = 0$. This means the graph should start at the point $(T_L, 0)$.

As $T_H$ increases, the efficiency increases. 

However, the efficiency can never exceed 1. As $T_H$ approaches infinity, the efficiency approaches 1. 

Therefore, the graph should asymptotically approach $\eta = 1$.

Rearranging the efficiency equation, we get:

$\eta = 1-\frac{T_L}{T_H} $ $1 - \eta = \frac{T_L}{T_H} $ $T_H = \frac{T_L}{1-\eta}$

The correct answer is (B)