• # question_answer A cannot engine working between 300 K and 600 K has a work output of 800 J/cycle. The amount of heat energy supplied to the engine from the source in each cycle is A)  800 J             B)  1600 JC)  3200 J                       D)  6400 J

Efficiency$=\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}=1-\frac{300}{600}=0.5$ $\eta =\frac{\text{Work}\,\text{done}}{\text{Heat}\,\text{supplied}}=\frac{W}{Q}$ Therefore, $\frac{1}{2}=\frac{800}{Q},\,\therefore \,Q=1600\,J$ Hence, the correction option is [b].