The temperatures of source and sink of a heat engine are $${127^ \circ }C$$  and $${27^ \circ }C$$  respectively. An inventor claims its efficiency to be $$26\% ,$$  then,

Solution :
Efficiency of heat engine is, $$\eta = 1 - \frac{{{T_2}}}{{{T_1}}}\,\,{\text{or}}\,\,\eta = \frac{{{T_1} - {T_2}}}{{{T_1}}}$$
$${{T_2}} =$$  temperature of sink
$${{T_1}} =$$  temperature of source
Given, $${T_1} = 273 + 127 = 400\,K$$
$${T_2} = 273 + 27 = 300\,K$$
$$\therefore \eta = \frac{{400 - 300}}{{400}} = \frac{{100}}{{400}} = 0.25 = 25\% $$
Hence, $$26\% $$  efficiency is impossible for a given heat engine.