JEE Main 2020 · 4 September · morning · Single correct · +4 / −1

The specific heat of water
= 4200 J kg⁻¹K⁻¹ and the latent heat of
ice = 3.4 × 10⁵ J kg –1 . 100 grams of ice at
0ᵒC is placed in 200 g of water at 25ᵒC. The
amount of ice that will melt as the temperature
of water reaches 0ᵒC is close to (in grams) :

  1. (a) 63.8
  2. (b) 61.7
  3. (c) 69.3
  4. (d) 64.6
Answer and solution

Answer (b) 61.7

Solution

Physics sir · 4 steps

Heat given out by the water cooling to 0 °C is exactly what melts the ice, so set the two equal.

  1. Heat from water. The 200 g of water cools from 25 °C to 0 °C. Use m = 0.2 kg, so the heat it gives out is Q = mc Δ T = 0.2 × 4200 × 25 = 21000 J.
  2. Heat to melt ice. The ice is already at 0 °C, so all of this heat goes into melting. Melting a mass m_(ice) takes m_(ice) L, with no change in temperature.

    m_(ice) = Q/L = 21000/(3.4 × 10⁵) = 0.0618 kg

  3. Check the ice supply. About 61.8 g needs to melt, and 100 g of ice is present. So the ice does not run out, and the mixture settles at 0 °C with some ice left over.
  4. So. About 61.7 g of ice melts, which is option (b).

The idea to keep. In ice and water problems, first find how much heat the warm water can give before it reaches 0 °C. Then compare it with the heat needed to melt all the ice to see whether any ice is left.

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

Study Thermometry & Calorimetry

Asked about 50 times in JEE Main since 2020. This question is from 2020.