In an insulated vessel, 0.05 kg steam at 373 K and 0.45 kg of ice at 253 K are mixed. Then, find the final temperature of the mixture. Given, \(L_{\text{fusion}}=80\) cal/g\(=336\) J/g, \(L_{\text{vaporization}}=540\) cal/g\(=2268\) J/g, \(S_{\text{ice}}=2100\) J/kg K\(=0.5\) cal/gK and \(S_{\text{water}}=4200\) J/kg K\(=1\) cal/gK

JEE Advanced · 2006 · 09 Apr 2006 · Heat & Thermodynamics

In an insulated vessel, 0.05 kg steam at 373 K and 0.45 kg of ice at 253 K are mixed. Then, find the final temperature of the mixture.

Given,

L_{fusion} = 80

cal/g

= 336

J/g,

L_{vaporization} = 540

cal/g

= 2268

J/g,

S_{ice} = 2100

J/kg K

= 0.5

cal/gK and

S_{water} = 4200

J/kg K

= 1

cal/gK

Answer: 273

Solution

Using the principle

ΣΔ Q = 0

:

Heat lost by steam to convert to 0°C water:

H_{L} = m_{s t e am} \times L_{v a p} + m_{s t e am} \times c_{w a t er} \times Δ T

H_{L} = 0.05 \times 540 + 0.05 \times 1 \times 100 = 27 + 5 = 32

kcal

Heat required by ice to convert to 0°C water:

First, heat ice from 253 K to 273 K:

Q_{1} = 0.45 \times 0.5 \times 20 = 4.5

kcal

Then, melt ice at 273 K:

Q_{2} = 0.45 \times 80 = 36

kcal

Total heat required:

H_{g} = 4.5 + 36 = 40.5

kcal

Since

H_{L} = 32

kcal

< H_{g} = 40.5

kcal, all the steam condenses and cools to 0°C, but there is insufficient heat to melt all the ice. The final state is a mixture of ice and water at 0°C.

Final temperature = 273 K (or 0°C)

JEE Advanced · 2006 · 09 Apr 2006 · Numerical answer

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JEE Advanced · 2006 · 09 Apr 2006 · Heat & Thermodynamics

Solution

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