106amount of heat liberated due to friction = change in KE of body = 60*100/2 = 3000 J
heat absorbed to melt the ice = 0.8*3000 = 2400J = 2400/4.2 cal
amt of ice melted = Q/L = (2400/4.2)/80 gm
= 7.1 gm
An ice-skater, weighing 60 kg moving at 10 m/s glides to a stop. If 80% of the heat generated by friction is
absorbed by the ice, the amount of ice melted will be nearly
(1) 30 g (2) 14 g (3) 7.1 g (4) 4.1 g
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