a) For the system to cool down to 0 C, a certain amount of heat energy must be provided so that only a part of ice is melted and not the whole, otherwise the equilibrium will shift from 0 C
Let us calculate the energy required to melt whole of the ice
Q = m L where L is latent heat of fusion
= 150 gms × 80cal/gms =12000 cal
The energy supplied by water at max can be
Q = m c ∆T = 250×1×70 = 17,500 cal
So yes the water has sufficient energy, first to melt the ice and then raise the temperature of the system as a whole so final equilibrium temperature will not be 0°C
b) The final state of equilibrium will be purely water and no ice and the water in a beaker will be at a temperature greater than 0°C.
c) To calculate the final equilibrium temperature,
As we have seen, after converting whole of ice into water, 17500-12000 cal = 5500 cal will still be left.
So using heat equation.
Q = m s ∆T
5500 = 400 × 1 × ∆T ( m = 250 + 150= 400 as all ice has been converted into water)
∆T = 13.75 °C
Hence the equilibrium temperature of the system will be 13.75 °C with 400gm water and 0g ice.
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