When ‘m’ g of ice is added to ‘M’ g of water at 20 °C, state the conditions for m and M for which
i) temperature of the mixture remains 0° C.
ii) temperature of the mixture exceeds 0 °C.
(Specific heat of water = sw = 4.2 × 103 J kg-1 °C-1, latent heat of fusion = L = 3.36 × 105 J kg-1)
The heat lost by water in going from 20 °C to 0°C,
Q1 = Msw ∆T = \(\frac{M}{1000}\) × 4.2 × 103 × (20) = 84M J
Now, heat required to convert m g of ice into water at 0 °C,
Q2 = mL = \(\frac{M}{1000}\) × 3.36 × 105 = 336m J
1. For temperature of mixture to be 0 °C,
Q2 > Q1
⇒ 336m > 84M
⇒ m > \(\frac{M}{4}\)
2. For temperature of mixture to exceed 0 °C,
Q2 < Q1 ⇒ m < \(\frac{M}{4}\)