Heat and Thermal ExpansionHard

Question

540 g of ice at 0oC is mixed with 540g of water at 80oC. The temperature of the mixture is (Give latent heat of fusion of ice = 80cal / g and specific heat capacity of water = 1 cal / goC)

Options

A.0oC
B.40oC
C.80oC
D.less than 0oC

Solution

Heat taken by ice to melt at 0oC is Q2 = mL = 540 × 80 = 43200 cal
Heat given by water to  cool 0oC is Q2 = msᐃθ 540 × 1 × (80 - 0) = 43200 cal
Hence heat given by water is just by water is just sufficient to melt the whole ice and final temperature of mixture is 0oC.

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