Question

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 = s_w = 4.2 × 10³ J kg^-1 °C^-1, latent heat of fusion = L = 3.36 × 10⁵ J kg^-1)

Answer 1

The heat lost by water in going from 20 °C to 0°C,

Q₁ = Ms_w ∆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 < \(\frac{M}{4}\)