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Heat and Thermodynamics question

2020 · 7 Jan · Shift 2 · Q41
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Heat and Thermodynamics question

2020 · 7 Jan · Shift 2 · Q41

JEE MainPhysicsHeat and ThermodynamicsNumerical+4 / −1
M grams of steam at 100oC is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produced liquid water at 40oC [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g] the value of M is ‾\underline{\hspace{2cm}}​
Numerical answer
View written solutionFree

Correct answer: 40

  1. Given data
  • Mass of steam =M= M=M g at 100∘C100^\circ\text{C}100∘C
  • Mass of ice =200= 200=200 g at 0∘C0^\circ\text{C}0∘C
  • Final state: liquid water at 40∘C40^\circ\text{C}40∘C
  • Latent heat of vaporization of water =540 cal/g= 540\ \text{cal/g}=540 cal/g
  • Latent heat of fusion of ice =80 cal/g= 80\ \text{cal/g}=80 cal/g
  • Specific heat of water =1 cal/g∘C= 1\ \text{cal/g}^\circ\text{C}=1 cal/g∘C

Since the container is thermally insulated,

Heat lost by steam=Heat gained by ice\text{Heat lost by steam} = \text{Heat gained by ice}Heat lost by steam=Heat gained by ice

  1. Heat lost by 1 g of steam

Each gram of steam at 100∘C100^\circ\text{C}100∘C first condenses to water at 100∘C100^\circ\text{C}100∘C, then cools to 40∘C40^\circ\text{C}40∘C.

  • Heat released during condensation: 540 cal/g540\ \text{cal/g}540 cal/g
  • Heat released during cooling from 100∘C100^\circ\text{C}100∘C to 40∘C40^\circ\text{C}40∘C: 1×(100−40)=60 cal/g1 \times (100-40) = 60\ \text{cal/g}1×(100−40)=60 cal/g

So, heat lost by MMM g steam is

M(540+60)=600MM(540+60)=600MM(540+60)=600M

  1. Heat gained by 200 g of ice

The ice at 0∘C0^\circ\text{C}0∘C first melts, then the melted water is heated to 40∘C40^\circ\text{C}40∘C.

  • Heat required to melt 200200200 g ice: 200×80=16000 cal200 \times 80 = 16000\ \text{cal}200×80=16000 cal
  • Heat required to raise temperature of 200200200 g water from 0∘C0^\circ\text{C}0∘C to 40∘C40^\circ\text{C}40∘C: 200×1×40=8000 cal200 \times 1 \times 40 = 8000\ \text{cal}200×1×40=8000 cal

Total heat gained:

16000+8000=24000 cal16000+8000=24000\ \text{cal}16000+8000=24000 cal

  1. Apply heat balance

600M=24000600M = 24000600M=24000

M=24000600=40M = \frac{24000}{600} = 40M=60024000​=40

  1. Final answer

40\boxed{40}40​

The derived answer matches the stored correct answer.

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