question_answer 12)

                  If an average person jogs, he produces \[14.5\times {{10}^{3}}\,\,cal/\min \]. This is removed by the evaporation of sweat. The amount of sweat evaporated of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires \[\text{58}.0\times \text{1}{{0}^{3}}\] cal for evaporation) is                 (a) 0.25 kg                           (b) 2.25 kg                 (c) 0.05 kg                            (d) 0.20 kg                

Answer:

                  (a) Amount of sweat evaporated \[=\,\frac{14.5\,\times \,{{10}^{3}}}{58\times \,{{10}^{3}}}=0.25\,kg\].