InterviewSolution
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InterviewSolution
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`4 gm` of steam at `100^(@)C` is added to `20 gm` of water at `46^(@)C` in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation `= 540 cal//gm`. Specific heat of water `=1 cal//gm-^(@)C`.A. `18 gm`B. `20 gm`C. `22 gm`D. `24 gm` |
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Answer» Correct Answer - C Heat released by steam in conversion to water a `100^(@)C` is `Q_(1)=mL=4xx540=2160 cal`. Heat required to raise temperature of water from `46^(@)C` to `100^(@)C` is `Q_(2) = mS Delta theta = 20xx1xx54=1080J` `Q_(1) gt Q_(2) and (Q_1)/(Q_2) =2` Hence all steam is not converted to water only half steam shall be converted to water Final mass of water =`20+2=22gm`. |
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