InterviewSolution
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InterviewSolution
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At `100^(@)C` the volume of `1 kg` of water is `10^(-3) m^(3)` and volume of `1 kg` of steam at normal pressure is `1.671 m^(3)`. The latent heat of steam is `2.3 xx 10^(6) J//kg` and the normal pressure is `10^(5) N//m^(2)`. If `5 kg` of water at `100^(@)C` is converted into steam, the increase in the internal energy of water in this process will beA. `8.35 xx 10^(5) J`B. `10.66 xx 10^(6) J`C. `11.5 xx 10^(5) J`D. zero |
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Answer» Correct Answer - B b. Heat required to convert 5 kg of water into steam `Delta Q = mL = 5 xx 2.3 xx 10^(6) = 11.10^(6) J` Work done in expanding volume, `Delta W = P Delta V` `= 5 xx 10^(5) (1.671 - 10^(-3)) = 0.835 xx 10^(6) J` Now by the first law of thermodynamics `Delta U = Delta Q - Delta W` `implies Delta U = 11.5 xx 10^(6) - 0.835 xx 10^(6) J` |
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