Kerala KEAM 5 June 2024 Solved Paper

The efficiency $\eta$ of a Carnot engine is given by the following equation:$$\eta = 1 - \frac{T_{sink}}{T_{source}}$$where:- $\eta$ is the efficiency of the engine,- $T_{sink}$ is the temperature of the sink (cold reservoir),- $T_{source}$ is the temperature of the source (hot reservoir).We are given:- $\eta = 50\% = 0.5$,- $T_{sink} = 350 \, K$.Substitute these values into the efficiency equation:$$0.5 = 1 - \frac{350}{T_{source}}$$Solving for $T_{source}$:$$\frac{350}{T_{source}} = 1 - 0.5 = 0.5$$$$T_{source} = \frac{350}{0.5} = 700 \, K$$Thus, the temperature of the source is $700 \, K$.$$\boxed{700 \, K}$$ Quick Tip: The efficiency of a Carnot engine is related to the temperatures of the hot and cold reservoirs. The formula $\eta = 1 - \frac{T_{sink}}{T_{source}}$ helps calculate the required temperature of the source given the efficiency and temperature of the sink.