First slide

Questions based on phase change

Question

60 g of ice at 0⁰C is mixed with 60 g of steam at 100⁰C. At thermal eqilibrium, the mixture contains

Moderate

A

80 g of water and 40 g of steam at 100⁰C
B

120 g of water 90⁰C
C

120 g of water at 100⁰C
D

40 g of steam and 80 g of water at 0⁰C

Solution

mL_S = Q₁
Q₁ = 60 x 540 = 32400 → steam to water at 100⁰C
mLi + msΔt = Q₂
Q₂ = 60 x 80 + 60 x 1 x 100 = 10800 → ice to water at 100⁰C
Since Q₁ > Q₂,a part of the steam is condensed.
If x gm of steam is converted into water at 100⁰C
$\large \therefore x\times 540=10800\Rightarrow x=20gm$
$\large \therefore$ so the final mixture contains (60 - 20)gm of steam is 40gm of steam and (60 + 20) gm of water i.e. 80gm of water.

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