A glass contains 650 g of water at 25 °C. 150 g of ice at -5 °C is put into the water. Calculate the final temperature of the contents.
Specific heat capacity of ice = 2030 J/kg K
Latent heat of fusion of water = 3.36 * 105 J/kg
Specific heat capacity of water = 4200 J/kg K
Latent heat of vaporisation of water = 2.26 * 106 J/kg
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Answer:
Energy required for ice from -5 °C to 0 °C
Q = mcΔθ = 0.15 * 2030 * 5
= 1522.5 J
Energy required for ice to convert from ice at 0 °C to water at °C
Q = ml = 0.15 * 336000
= 50400 J
Thus, total heat gained by ice = 51922.5 J
Note: This heat gained by ice (heat loss by the water) is not yet sufficient to bring the water to 0 °C and convert it to ice. Hence, we work assuming that the final temperature is somewhere between 25 °C and 0 °C, which we shall see.
Let the final temperature be t
total heat loss by water = total heat gained by ice
mcwaterΔθ (from water at 25 °C to water at t °C) = 51922.5 + mcwaterΔθ (from water at 0 °C to water at t °C)
0.65 * 4200 * (25 - t) = 51922.5 + 0.15 * 4200 * t
68250 - 2730 t = 51922.5 + 630 t
3360 t = 16327.5
t = 4.86 °C
