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JEE Main 201912 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
12 April
Shift
Morning Shift
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Main 2019 · 12 April · Morning Shift · Physics · Thermal Properties of Matter

When M _ 1 gram of ice at - 10 ^ o C (specific heat = 0.5cal g ^ -...

When M1{M}_{1} gram of ice at −10oC- {10}^{o}C (specific heat =0.5calg−1℃−1= 0.5cal{g}^{- 1}{℃}^{- 1} ) is added to M2{M}_{2} gram of water at 50oC,{50}^{o}C, finally no ice is left and the water is at 0oC{0}^{o}C . The value of latent heat of ice, in calg−1cal{g}^{- 1} is:

Options

  1. A50M2M1\dfrac{50{M}_{2}}{{M}_{1}}
  2. B5M1M2−50\dfrac{5{M}_{1}}{{M}_{2}} - 50
  3. C5M2M1−5\dfrac{5{M}_{2}}{{M}_{1}} - 5
  4. D50M2M1−5\dfrac{5{0M}_{2}}{{M}_{1}} - 5

Correct answer

D. 50M2M1−5\dfrac{5{0M}_{2}}{{M}_{1}} - 5

Step-by-step solution

Principle of calorimetry
M1Sice(0−(−10))+M1Lf=M2Swater(50−0){M}_{1}{S}_{ice}\left(0 - \left(- 10\right)\right) + {M}_{1}{L}_{f} = {M}_{2}{S}_{water}\left(50 - 0\right)
M1(0.5)(10)+M1Lf=M2(50){M}_{1}\left(0.5\right)\left(10\right) + {M}_{1}{L}_{f} = {M}_{2}\left(50\right)
5M1+M1Lf=M2(50)5{M}_{1} + {M}_{1}{L}_{f} = {M}_{2}\left(50\right)
Lf=50M2M1−5{L}_{f} = \dfrac{50{M}_{2}}{{M}_{1}} - 5
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