CalorimetryEasy
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.
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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