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Question

1 kg of ice at 0C is mixed with 1 kg of steam at 100C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 × 105 J kg1, specific heat capacity of water = 4200 J kg −1 K−1 and latent heat of vaporization of water = 2.26 × 106 J kg1


A

2kg steam

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B

2 kg ice

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C

665gm water, 1.335 kg steam

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D

665 gm steam, 1.335 kg water

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Solution

The correct option is D

665 gm steam, 1.335 kg water


As the mass of ice and steam is the same, let us start by assuming that all the ice converts to water at 0°C, then heats up to water at 100C, in the process converting some of the steam to water at 100C. Don't worry, if our assumption is incorrect, the mass of water will come out negative and we can then consider other possibilities!
So, Heat gained during ice at 0C water at 100C

Q1 = mL1 + msΔT

m = 1kg

L1 Latent heat of fusion
ΔT = 100C
S specific heat of water
Q1 = L1 + 100S

Now, heat released by x kg steam converted to water at is

Q2 = xL2

Where, L2 latent heat of vaporization

But, Q1 = Q2

L1 + 100 S = xL2
x = L1 + 100sL2

x = 0.335 kg steam converts to water

Amount of steam left = (1 0.335)kg = 665gm


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