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At normal pressure and 0^(@)C temperatur...

At normal pressure and `0^(@)C` temperature the volume of `1` kg of ice is reduced by `91 cm^(3)` on melting . Laters heat of melting of ice is `3.4xx 10^(5)J//kg`. Calculate the changes in the internal energy when `2`kg of ice melts normal pressure and `0^(@)C. (P=1.01 xx 10^(5) Nm^(-2))`

Text Solution

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Heat energy absorbed by `2` kg of ice melting `Q= "m"L = 2xx3.4xx10^(5) = 6.8 xx 10^(5)J`
Change in volume of `2` kg of ice = `2xx91 = 182cm^(3) = 182xx 10^(-6)m^(3)`
`therefore` `" "` `W= PDelta V = 1.01 xx 10^(5) xx (-182xx10^(-6)) = -18.4J`
Since, work is done on ice so work `W` is taken -ve . Now from first law of thermodynamics
` Q = DeltaU + W implies Delta U = Q-W = 6.8 xx 10^(5)- (-18.4) = (6.8xx10^(5) + 18.4)J = 680018.4J`
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