question_answer

If you are asked to increase the efficiency of a Carnot engine by increasing the temperature of the source or by decreasing the temperature of the sink by\[\mathbf{10}\text{ }\mathbf{K}\], which method would you prefer and why?

Answer:

                Let \[{{T}_{1}}\] and \[{{T}_{2}}\] be initial temperatures of source and sink respectively. When temperature of the source is increased by\[\text{1}0\text{ K}\], the efficiency becomes \[{{\eta }_{1}}=1-\frac{{{T}_{2}}}{{{T}_{1}}+10}\] When the temperature of the sink is decreased by \[\text{1}0\text{ K},\] the efficiency becomes \[{{\eta }_{2}}=1-\frac{{{T}_{2}}-10}{{{T}_{1}}}\] It can be easily seen that \[{{\eta }_{2}}-{{\eta }_{1}}>0\] or \[{{\eta }_{2}}>{{\eta }_{1}}\] Hence the efficiency of a Carnot engine can be, increased by a greater amount by decreasing the temperature of the sink by \[\text{1}0\text{ K}\] than by increasing the temperature of the source by\[\text{1}0\text{ K}\].