A calorimeter of water equivalent 20 g contains 180 g of water at $25^{\circ} \mathrm{C}$, 'm' grams of steam at $100^{\circ} \mathrm{C}$ is mixed in it till the temperature of the mixture is $31^{\circ} \mathrm{C}$. The value of 'm' is close to (Latent heat of water $=540 \mathrm{cal} \mathrm{g}^{-1}$, specific heat of water $=1 \mathrm{calg}^{-1}{ }^{\circ} \mathrm{C}^{-1}$)

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Heat & Thermodynamics

A calorimeter of water equivalent 20 g contains 180 g of water at

2 5^{\circ} C

, 'm' grams of steam at

10 0^{\circ} C

is mixed in it till the temperature of the mixture is

3 1^{\circ} C

. The value of 'm' is close to (Latent heat of water

= 540 cal g^{- 1}

, specific heat of water

= 1 calg^{- 1}^{\circ} C^{- 1}

)

(A)2.6
(B)4
(C)2✓ Correct
(D)3.2

Answer: (C)

Solution

Equate the heat released by the steam to the heat absorbed by the water and the calorimeter. The steam first condenses, giving up its latent heat, and the resulting water then cools from

10 0^{\circ}

C to

3 1^{\circ}

C, so

m \times 540 + m \times 1 \times (100 - 31) = (180 + 20) \times 1 \times (31 - 25)

. This becomes

540 m + 69 m = 200 \times 6

, that is

609 m = 1200

, so

m \approx 2 g

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option

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