A Carnot engine operating between two reservoirs has efficiency $\frac{1}{3}$ When the temperature of cold reservoir raised by $x$, its efficiency decreases to $\frac{1}{6}$ The value of $x$, if the temperature of hot reservoir is $99^{\circ} C$, will be :

Question

A Carnot engine operates between temperatures of $ 600 \, \text{K} $ and $ 300 \, \text{K} $. If it absorbs $ 900 \, \text{J} $ of heat from the source, how much work does it perform?

Answer 1

To solve this problem, we will use the formula for the efficiency of a Carnot engine. The efficiency ($\eta$) of a Carnot engine operating between two temperatures $T_h$ (hot reservoir) and $T_c$ (cold reservoir) is given by:

$\eta = 1 - \frac{T_c}{T_h}$

Here, the temperatures should be in Kelvin. Let's start with the initial setup:

The initial efficiency is given as $\frac{1}{3}$. Therefore:

$\frac{1}{3} = 1 - \frac{T_{c1}}{T_{h}}$

Simplifying, we get:

$\frac{T_{c1}}{T_{h}} = \frac{2}{3}$

This implies:

$T_{c1} = \frac{2}{3}T_{h}$

When the temperature of the cold reservoir is raised by $x$, the efficiency becomes $\frac{1}{6}$. Therefore:

$\frac{1}{6} = 1 - \frac{T_{c2}}{T_{h}}$

Simplifying, we get:

$\frac{T_{c2}}{T_{h}} = \frac{5}{6}$

This implies:

$T_{c2} = \frac{5}{6}T_{h}$

The increase in the temperature of the cold reservoir is given by:

$T_{c2} = T_{c1} + x$

From the equations derived above:

$\frac{5}{6}T_{h} = \frac{2}{3}T_{h} + x$

Rearranging for $x$:

$x = \frac{5}{6}T_{h} - \frac{2}{3}T_{h}$

Because $\frac{2}{3} = \frac{4}{6}$, we find:

$x = \left(\frac{5}{6} - \frac{4}{6}\right)T_{h}$

$x = \frac{1}{6}T_{h}$

Now, substitute $T_{h}$ with the given temperature of the hot reservoir. Converting the given temperature in Celsius to Kelvin:

$T_{h} = 99^\circ C + 273 = 372K$

Then,

$x = \frac{1}{6} \times 372$

Calculating gives:

$x = 62K$

Therefore, the value of $x$ is $62 K$. The correct answer is $62 K$.

Answer 2

To solve this problem, we will use the formula for the efficiency of a Carnot engine. The efficiency ($\eta$) of a Carnot engine operating between two temperatures $T_h$ (hot reservoir) and $T_c$ (cold reservoir) is given by:

$\eta = 1 - \frac{T_c}{T_h}$

Here, the temperatures should be in Kelvin. Let's start with the initial setup:

The initial efficiency is given as $\frac{1}{3}$. Therefore:

$\frac{1}{3} = 1 - \frac{T_{c1}}{T_{h}}$

Simplifying, we get:

$\frac{T_{c1}}{T_{h}} = \frac{2}{3}$

This implies:

$T_{c1} = \frac{2}{3}T_{h}$

When the temperature of the cold reservoir is raised by $x$, the efficiency becomes $\frac{1}{6}$. Therefore:

$\frac{1}{6} = 1 - \frac{T_{c2}}{T_{h}}$

Simplifying, we get:

$\frac{T_{c2}}{T_{h}} = \frac{5}{6}$

This implies:

$T_{c2} = \frac{5}{6}T_{h}$

The increase in the temperature of the cold reservoir is given by:

$T_{c2} = T_{c1} + x$

From the equations derived above:

$\frac{5}{6}T_{h} = \frac{2}{3}T_{h} + x$

Rearranging for $x$:

$x = \frac{5}{6}T_{h} - \frac{2}{3}T_{h}$

Because $\frac{2}{3} = \frac{4}{6}$, we find:

$x = \left(\frac{5}{6} - \frac{4}{6}\right)T_{h}$

$x = \frac{1}{6}T_{h}$

Now, substitute $T_{h}$ with the given temperature of the hot reservoir. Converting the given temperature in Celsius to Kelvin:

$T_{h} = 99^\circ C + 273 = 372K$

Then,

$x = \frac{1}{6} \times 372$

Calculating gives:

$x = 62K$

Therefore, the value of $x$ is $62 K$. The correct answer is $62 K$.

A Carnot engine operating between two reservoirs has efficiency \(\frac{1}{3}\) When the temperature of cold reservoir raised by \(x\), its efficiency decreases to \(\frac{1}{6}\) The value of \(x\), if the temperature of hot reservoir is \(99^{\circ} C\), will be :

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The Correct Option is A

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