M grams of steam at 100ᵒC is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produced liquid water at 40ᵒC [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g] the value of M is ____

Solution

Physics sir · 3 steps

Steam gives out heat in two stages and ice takes in heat in two stages: set the two totals equal.

1. Heat given by steam. The steam first condenses at 100°C, giving 540M cal. That water then cools from 100°C to 40°C, giving M × 1 × 60 = 60M cal. Total given out: 600M cal.
2. Heat taken by ice. The ice first melts at 0°C, taking 200 × 80 = 16000 cal. That water then warms from 0°C to 40°C, taking 200 × 1 × 40 = 8000 cal. Total taken in: 24000 cal.

   M(540 + 60) = 200(80 + 40) ⇒ 600M = 24000

3. So. The container is insulated, so the heat given out equals the heat taken in. This gives M = 24000/600 = 40 g.

The common slip: 44.4. Leaving out the cooling of the condensed water from 100°C to 40°C gives 540M = 24000, so M ≈ 44.4.

The idea to keep. In mixing problems, trace each substance from its start state to the final temperature. Count every phase change and every temperature change, then equate the heat lost to the heat gained.

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