A sample of 0.1 text{ g} of water at 100^{cir | vedclass

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(B) Given: Mass $m = 0.1 	ext{ g} = 10^{-4} 	ext{ kg}$. Heat supplied $Q = 54 	ext{ cal} = 54 	imes 4.184 	ext{ J} \approx 225.94 	ext{ J}$. Pre

Vedclass · Accepted Answer

$208.7$

Vedclass · Answer

(B) Given: Mass $m = 0.1 	ext{ g} = 10^{-4} 	ext{ kg}$. Heat supplied $Q = 54 	ext{ cal} = 54 	imes 4.184 	ext{ J} \approx 225.94 	ext{ J}$. Pressure $P = 1.013 	imes 10^5 	ext{ Pa}$. Volume of water $V_w = 0.1 	ext{ cc} = 10^{-7} 	ext{ m}^3$. Volume of steam $V_s = 167.1 	ext{ cc} = 167.1 	imes 10^{-6} 	ext{ m}^3$. Change in volume $\Delta V = V_s - V_w = 167.1 	imes 10^{-6} - 0.1 	imes 10^{-6} = 167 	imes 10^{-6} 	ext{ m}^3$. Work done $W = P \Delta V = (1.013 	imes 10^5) 	imes (167 	imes 10^{-6}) \approx 16.92 	ext{ J}$. Using the first law of thermodynamics: $\Delta U = Q - W$. $\Delta U = 225.94 - 16.92 = 209.02 	ext{ J}$. Rounding to the nearest option,the value is approximately $208.7 	ext{ J}$.

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