A calorimeter of water equivalent 20 g contains 180 g of water at 25ᵒC. ‘m’ grams of steam at 100ᵒC is mixed in it till the temperature of the mixure is 31ᵒC. The value of ‘m’ is close to :
(Latent heat of water = 540 cal g –1 , specific heat of water = 1 cal g –1 ᵒC –1 )

Solution

Physics sir · 4 steps

Steam gives heat in two stages: first it condenses, then that hot water cools down to 31°C.

1. Heat gained. The water equivalent of 20 g means the calorimeter behaves like 20 g of extra water. So 20 + 180 = 200 g of water warms from 25°C to 31°C and gains 200 × 1 × 6 = 1200 cal.
2. Heat lost. Each gram of steam at 100°C condenses and gives 540 cal. That water then cools from 100°C to 31°C and gives 1 × 69 = 69 cal more. So the steam loses 609m cal in total.

   m × 540 + m × 1 × (100 - 31) = 200 × 1 × (31 - 25)

3. Solve for m. 609m = 1200, so m = 1200/609 ≈ 1.97 g.
4. So. m is close to 2 g, which is option (b).

The idea to keep. When steam is mixed in, count both the latent heat and the cooling of the condensed water down to the final temperature. Add the water equivalent to the water mass before you multiply.

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