1;g of water,of volume 1;cm^{3} at 100^{circ} | vedclass

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(B) According to the first law of thermodynamics,$Q = \Delta U + W$.Here,$Q$ is the heat supplied,$\Delta U$ is the change in internal energy,and $W$

Vedclass · Accepted Answer

$2089$

Vedclass · Answer

(B) According to the first law of thermodynamics,$Q = \Delta U + W$.Here,$Q$ is the heat supplied,$\Delta U$ is the change in internal energy,and $W$ is the work done by the system.Given:Mass $m = 1\;g$Latent heat $L = 2256\;J/g$Pressure $P = 1 	imes 10^{5}\;Pa$Initial volume $V_{1} = 1\;cm^{3} = 1 	imes 10^{-6}\;m^{3}$Final volume $V_{2} = 1671\;cm^{3} = 1671 	imes 10^{-6}\;m^{3}$Heat supplied $Q = mL = 1\;g 	imes 2256\;J/g = 2256\;J$.Work done $W = P(V_{2} - V_{1}) = 1 	imes 10^{5} 	imes (1671 - 1) 	imes 10^{-6} = 10^{5} 	imes 1670 	imes 10^{-6} = 167\;J$.Using $Q = \Delta U + W$,we get $\Delta U = Q - W = 2256 - 167 = 2089\;J$.

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