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A `50 kg` man is running at a speed of `18 kmH^(-1)` If all the kinetic energy of the man be uses to increase the temperature of water from `30^(@)C`how much water can be beated with this energy? |
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Answer» Correct Answer - A K.E. of the mass `== (1)/(2) mV^(2)` `= ((1)/(2)) 50 xx 5^(2)` `= 25 xx 25 625J` The amount of heat required to raise the temperature of water from `20^(@)C to 30^(@)C ` `msDelta theta= m xx(30 - 20)` `4200m` `But, `42 xx 10^(3) m = 625` `rArr m = (625)/(42) xx 10^(-3)` ` = 14.88 xx 10^(-3) kg` `15 g` |
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