Q.Consider a Carnot's cycle operating between $T_1 = 500$ K and $T_2 = 300$ K producing 1 kJ of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.

The key idea here is the efficiency of a Carnot engine, which sets the theoretical maximum efficiency for any heat engine operating between two given temperatures.

1. The efficiency ($\eta$) of a Carnot engine is determined by the temperatures of the hot reservoir ($T_1$) and the cold reservoir ($T_2$):

2. Efficiency is also defined as the ratio of the mechanical work done ($W$) to the heat absorbed from the hot reservoir ($Q_1$): $\eta = \frac{W}{Q_1}$ …

Concept and Intuition: The Carnot Cycle and its Efficiency

A heat engine is a device that converts thermal energy (heat) into mechanical energy (work). The Carnot cycle represents the most efficient possible cycle for a heat engine operating between two given temperature reservoirs. This theoretical limit is crucial because it tells us the maximum fraction of heat input that can be converted into useful work.

The efficiency of any heat engine ($\eta$) is defined as the ratio of the useful work output ($W$) to the heat absorbed from the hot reservoir ($Q_1$).

For a Carnot engine, this efficiency depends only on the absolute temperatures of the hot reservoir ($T_1$) and the cold reservoir ($T_2$). It is given by:

This formula is fundamental because it shows that to achieve higher efficiency, the temperature difference between the hot and cold reservoirs must be as large as possible. An efficiency of 1 (100%) would only be possible if $T_2 = 0$ K, which is practically impossible.

The problem asks for the heat transferred to the engine ($Q_1$). We are given the work done ($W$) and the temperatures of the reservoirs ($T_1$ and $T_2$). By first calculating the Carnot efficiency using the temperatures, we can then use the definition of efficiency ($\eta = W/Q_1$) to find $Q_1$. This approach works because the Carnot efficiency directly links the work output, heat input, and the operating temperatures.

1. Identify Given Parameters:

   We are given the following information:

   • Temperature of the hot reservoir, $T_1 = 500$ K.
   • Temperature of the cold reservoir, $T_2 = 300$ K.
   • Mechanical work produced per cycle, $W = 1$ kJ.

2. Determine the Goal:

   We need to find the heat transferred to the engine by the hot reservoir, which is $Q_1$.

3. Calculate the Carnot Efficiency:

   The efficiency of a Carnot engine is given by the formula: …

Shortcut — use the Carnot heat ratio directly. Rather than computing the efficiency first and then inverting it, use $\dfrac{Q_1}{Q_2}=\dfrac{T_1}{T_2}$ together with $W=Q_1-Q_2$ for a Carnot cycle. Substituting $Q_2=Q_1T_2/T_1$ gives $W=Q_1\left(1-\dfrac{T_2}{T_1}\right)$, so $Q_1=\dfrac{W}{1-T_2/T_1}=\dfrac{WT_1}{T_1-T_2}=\dfrac{1\times500}{200}= …