A sample of 0.1 , g of water at 100^{circ} C | vedclass

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Question

(B) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.First,convert the heat energy from calories to Joules: $\Delta Q = 5

Vedclass · Accepted Answer

$208.7$

Vedclass · Answer

(B) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.First,convert the heat energy from calories to Joules: $\Delta Q = 54 \, cal 	imes 4.18 \, J/cal = 225.72 \, J$.Next,calculate the work done during expansion: $\Delta W = P \Delta V$.Given $P = 1.013 	imes 10^5 \, N/m^2$ and $\Delta V = 167.1 \, cc = 167.1 	imes 10^{-6} \, m^3$.$\Delta W = 1.013 	imes 10^5 	imes 167.1 	imes 10^{-6} \approx 16.93 \, J$.Now,calculate the change in internal energy: $\Delta U = \Delta Q - \Delta W$.$\Delta U = 225.72 \, J - 16.93 \, J = 208.79 \, J$.Rounding to the nearest option,$\Delta U \approx 208.7 \, J$.

$A$ sample of $0.1 \, g$ of water at $100^{\circ} C$ and normal pressure $(1.013 \times 10^5 \, N m^{-2})$ requires $54 \, cal$ of heat energy to convert to steam at $100^{\circ} C$. If the volume of the steam produced is $167.1 \, cc$,the change in internal energy of the sample is ....... $J$.

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