Question

when 50 g of ice at 0 degree c is placed in 400 g of water at 55 degree c, the final temperature is 40 degree c. if we ignore any exchange of heat to the surroundings, how much heat does it take to melt 1 g of ice at 0 degree c

Answer #1

Let heat required to melt 1 gm ice be L cal/gm. Heat required to melt 50 gm ice = 50L cal.

Heat required to increase temperature of 50 gm water at 0 deg C (after melting) to 40 deg C

= (mass)*(specific heat of water)*(change in temperature)

= (50 gm)*(1cal/gm C)*(40 C - 0 C) = 2000 cal

Heat transferred by 400 gm hot water at 55 deg C to 40 deg C

= (400 gm)*(1 cal/gm C)*(55 C - 40 C) = 6000 cal

Heat received by ice is equal to heat given by hot water. Hence

50 L + 2000 = 6000

L = 80 cal/gm

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