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Question
Four 2 cm × 2 cm × 2 cm cubes of ice are taken out from a refrigerator and are put in 200 ml of a drink at 10°C. (a) Find the temperature of the drink when thermal equilibrium is attained in it. (b) If the ice cubes do not melt completely, find the amount melted. Assume that no heat is lost to the outside of the drink and that the container has negligible heat capacity. Density of ice = 900 kg m−3, density of the drink = 1000 kg m−3, specific heat capacity of the drink = 4200 J kg−1 K−1, latent heat of fusion of ice = 3.4 × 105 J kg−1.
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Solution
(a)
Given:-
Number of ice cubes = 4
Volume of each ice cube = (2 × 2 × 2) = 8 cm3
Density of ice = 900 kg m−3
Total mass of ice, mi = (4 × 8 ×10−6 ×900) = 288×10−4 kg
Latent heat of fusion of ice, Li = 3.4 × 105 J kg−1
Density of the drink = 1000 kg m−3
Volume of the drink = 200 ml
Mass of the drink = (200×10−6)×1000 kg
Let us first check the heat released when temperature of 200 ml changes from 10oC to 0oC.
Hw = (200×10−6)×1000×4200×(10−0) = 8400 J
Heat required to change four 8 cm3 ice cubes into water (Hi) = miLi = (288×10−4)×(3.4×105) = 9792 J
Since the heat required for melting the four cubes of the ice is greater than the heat released by water ( Hi > Hw ), some ice will remain solid and there will be equilibrium between
ice and water. Thus, the thermal equilibrium will be attained at 0o C.
(b)
Equilibrium temperature of the cube and the drink = 0°C
Let M be the mass of melted ice.
Heat released when temperature of 200 ml changes from 10oC to 0oC is given by
Hw = (200×10−6)×1000×4200×(10−0) = 8400 J
Thus,
M×(3.4×105) = 8400 J
Therefore,
M = 0.0247 Kg = 25 g
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