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प्रश्न
उत्तर
Given, Sp. heat capacity of water, s = 4200 J/KgK
Temperature difference, Δt = 0.21 °C
Let 'h' be the height of the waterfall and 'm' be the mass of water.
Then, P.E of the water = mgh
Given that, heat energy, H = 60% of P.E
or, 4200 x 0.21 = 0.6 x 10 x h
or, h = `(4200 xx 0.21)/(0.6 xx 10) = 882/6` = 147 m
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