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An athlete is given 100g of glucose (C_(6)H_(12)O_(6)) of energy equivalent to 1560 kJ. He utilizes 50% of this gained energy in the event . In order to avoid storage of energy in the body, calculate the weight of water he would need to perspire. The enthalpy of evaporation of water is 44 "kJ mole"^(-1) |
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Answer» Solution :Energy remained in the body of the athlete after the event `=(1560)/(2)=780"kJ"` `:.` weight of water to be evaporated by `780 kJ` of energy `=(18)/(44)xx780=319.1g. (H_(2)O=18)` |
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