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If an avarage person jogs, he produces `14.5xx10^(4) cal//min`. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming `1 kg` requites `580xx10^(3)` cal for evaporation) isA. 0.25kgB. 2.25kgC. 0.05kgD. 0.20kg |
Answer» Correct Answer - A Amount of sweat evaprated/minute `("Sweat produced"//"minute")/("Number of calories required for evaporation"//kg)` `("Amount of heat produced per minute in jogging")/("Latenheat" ("in cal"//kg))` `(14.5xx10^(3))/(580xx10^(3))=(145)/(580)=0.25kg` |
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