Work done by a Carnot engine operating between temperatures $127^{\circ} \mathrm{C}$ and $27^{\circ} \mathrm{C}$ is $2 \mathrm{~kJ}$. The amount of heat transferred to the engine by the reservoir is :

<p>The efficiency of a Carnot engine is given by:</p> <p>$\eta = 1 - \frac{T_{c}}{T_{h}}$</p> <p>where:</p> <ul> <li>$\eta$ is the efficiency of the Carnot engine,</li> <li>$T_{c}$ is the temperature of the cold reservoir,</li> <li>$T_{h}$ is the temperature of the hot reservoir.</li> </ul> <p>Note that the temperatures must be in Kelvin for this formula.</p> <p>Here, the given temperatures are $127^{\circ}C$ and $27^{\circ}C$. Converting these to Kelvin gives:</p> <p>$T_{h} = 127^{\circ}C + 273.15 = 400.15 K$ $T_{c} = 27^{\circ}C + 273.15 = 300.15 K$</p> <p>So the efficiency of the Carnot engine is:</p> <p>$\eta = 1 - \frac{300.15}{400.15} = 0.25$</p> <p>The efficiency can also be defined as the work done divided by the heat supplied:</p> <p>$\eta = \frac{W}{Q_{h}}$</p> <p>where:</p> <ul> <li>$W$ is the work done by the engine,</li> <li>$Q_{h}$ is the heat transferred to the engine by the hot reservoir.</li> </ul> <p>We can rearrange this equation to solve for $Q_{h}$:</p> <p>$Q_{h} = \frac{W}{\eta} = \frac{2 \, kJ}{0.25} = 8 \, kJ$</p>