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Power of a man who can chew 0.3 kg ice in one minute

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2400 cal s^{- 1}

400 cal s^{- 1}

4 cal s^{- 1}

24 cal s^{- 1}

answer is A.

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Detailed Solution

m = 0.3 kg,t = 1 \min = 60 s

Power $= \frac{w}{t} = \frac{Q}{t} = \frac{mL}{t} = \frac{0.3 \times 80 \times 10^{3}}{60}$

$power = 400 cal/s$

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