A block of ice of mass 120 g at temperature 0°C is put in 300 g of water at 25°C. The x g of ice melts as the temperature of the water reaches 0°C. The value of x is _____________.
[Use specific heat capacity of water = 4200 Jkg⁻¹K⁻¹, Latent heat of ice = 3.5 × 10⁵ Jkg⁻¹]

Solution

Physics sir · 4 steps

Heat the water gives up cooling to 0°C goes entirely into melting ice, so equate the two.

1. Heat from water. The 300 g of water cools from 25°C to 0°C. Work in kilograms: Q = mcΔ T = 0.3 × 4200 × 25 = 31500 J.
2. Heat for ice. The ice is already at 0°C, so it needs no warming. Every joule goes into melting, which takes xL for a mass x.

   x = (m_w c Δ T)/L = 31500/(3.5 × 10⁵) = 0.09 kg

3. Check it fits. 0.09 kg = 90 g, which is less than the 120 g of ice available. So only part of the ice melts, and the mixture settles at 0°C with 30 g still frozen.
4. So. The mass of ice that melts is x = 90 g.

The common slip: 0.09. Leaving the answer in kilograms gives 0.09, but the blank asks for grams, so it is 90.

The idea to keep. In ice-water mixing, first find the heat the water can give up reaching 0°C. Compare it with the heat needed to melt all the ice: if it falls short, only part of the ice melts.

Answer from the NTA final answer key. Solution by PrepWiser.