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JEE Main 201323 AprilOnline
Exam
JEE Main
Year
2013
Date
23 April
Mode
Online
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Main 2013 · 23 April · Online · Physics · Thermal Properties of Matter

500 ~g of water and 100 ~g of ice at 0^ C are in a calorimeter...

500 g500 \mathrm{~g} of water and 100 g100 \mathrm{~g} of ice at 0∘C0^{\circ} \mathrm{C} are in a calorimeter whose water equivalent is 40 g.10 g40 \mathrm{~g} .10 \mathrm{~g} of steam at 100∘C100^{\circ} \mathrm{C} is added to it. Then water in the calorimeter is : (Latent heat of ice =80cal/g=80 \mathrm{cal} / \mathrm{g}, Latent heat of steam =540cal/g=540 \mathrm{cal} / \mathrm{g} )

Options

  1. A580 g580 \mathrm{~g}
  2. B590 g590 \mathrm{~g}
  3. C600 g600 \mathrm{~g}
  4. D610 g610 \mathrm{~g}

Correct answer

B. 590 g590 \mathrm{~g}

Step-by-step solution

As 1 g1 \mathrm{~g} of steam at 100∘C100^{\circ} \mathrm{C} melts 8 g8 \mathrm{~g} of ice at 0∘C0^{\circ} \mathrm{C}. 10 g10 \mathrm{~g} of steam will melt 8×10 g8 \times 10 \mathrm{~g} of ice at 0∘C0^{\circ} \mathrm{C} Water in calorimeter =500+80+10 g=590 g=500+80+10 \mathrm{~g}=590 \mathrm{~g}
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