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JIPMER 2015
Exam
JIPMER
Year
2015
Subject
Physics
Chapter
Thermodynamics

JIPMER 2015 · Physics · Thermodynamics

1 kg of water is converted into steam at the same temperature and...

1 kg of water is converted into steam at the same temperature and at 1 atm ( 100 kPa ). The density of water and steam are 1000kgm−31000 \mathrm{kgm}^{-3} and 0.6kgm−30.6 \mathrm{kgm}^{-3} respectively. The latent heat of vaporisation of water is 2.25×106Jkg−12.25 \times 10^6 \mathrm{Jkg}^{-1}. What will be increase in energy?

Options

  1. A3×105 J3 \times 10^5 \mathrm{~J}
  2. B4×106 J4 \times 10^6 \mathrm{~J}
  3. C2.08×106 J2.08 \times 10^6 \mathrm{~J}
  4. DNone of these

Correct answer

C. 2.08×106 J2.08 \times 10^6 \mathrm{~J}

Step-by-step solution

The volume of 1 kg water =1100 m3\frac{1}{100} \mathrm{~m}^3 and volumeof1kgsteam=10.6 m3volume of 1 kg steam =\frac{1}{0.6} \mathrm{~m}^3 The increase in volume =10.6 m3−11000 m3=(1.7−0.001)m3≅1.7 m3\begin{aligned} & =\frac{1}{0.6} \mathrm{~m}^3-\frac{1}{1000} \mathrm{~m}^3 \\ & =(1.7-0.001) \mathrm{m}^3 \\ & \cong 1.7 \mathrm{~m}^3\end{aligned} The work done by steam is W=(100kPa)(1.7 m3)=1.7×105 J\begin{aligned} W & =(100 \mathrm{kPa})\left(1.7 \mathrm{~m}^3\right) \\ & =1.7 \times 10^5 \mathrm{~J}\end{aligned} The change in intemal energy ΔU=ΔQ−ΔW=2.25×106 J−1.7×105 J=208×106 J\begin{aligned} \Delta U & =\Delta Q-\Delta W \\ & =2.25 \times 10^6 \mathrm{~J}-1.7 \times 10^5 \mathrm{~J} \\ & =208 \times 10^6 \mathrm{~J}\end{aligned}
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