1,g of water at 100,^{circ}C and 10^5,Pa pres | vedclass

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(C) Given: Mass $m = 1\,g = 10^{-3}\,kg$,Pressure $P = 10^5\,Pa$,Volume of steam $V_s = 1841\,cm^3 = 1841 	imes 10^{-6}\,m^3$,Latent heat $L = 2250\,

Vedclass · Accepted Answer

$2066$

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(C) Given: Mass $m = 1\,g = 10^{-3}\,kg$,Pressure $P = 10^5\,Pa$,Volume of steam $V_s = 1841\,cm^3 = 1841 	imes 10^{-6}\,m^3$,Latent heat $L = 2250\,J/g = 2250 	imes 10^3\,J/kg$.The volume of $1\,g$ of water is $V_w = \frac{m}{ho} = \frac{1\,g}{1\,g/cm^3} = 1\,cm^3 = 10^{-6}\,m^3$.According to the first law of thermodynamics,$\Delta Q = \Delta U + W$,where $W = P(V_s - V_w)$.Heat absorbed $\Delta Q = m 	imes L = 1\,g 	imes 2250\,J/g = 2250\,J$.Work done $W = P(V_s - V_w) = 10^5\,Pa 	imes (1841 	imes 10^{-6} - 1 	imes 10^{-6})\,m^3 = 10^5 	imes 1840 	imes 10^{-6} = 184\,J$.Change in internal energy $\Delta U = \Delta Q - W = 2250\,J - 184\,J = 2066\,J$.

$1\,g$ of water at $100\,^{\circ}C$ and $10^5\,Pa$ pressure converts into $1841\,cm^3$ of steam at constant temperature and pressure. If the latent heat of vaporization of water is $2250\,J/g$,the change in internal energy of water in this process is ..... $J$.

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