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

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(B) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$. Given: $\Delta Q = 54 \ cal = 54 	imes 4.184 \ J = 225.936 \ J$. W

Vedclass · Accepted Answer

$208.7$

Vedclass · Answer

(B) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$. Given: $\Delta Q = 54 \ cal = 54 	imes 4.184 \ J = 225.936 \ J$. Work done $\Delta W = P \Delta V = (1.013 	imes 10^5 \ N m^{-2}) 	imes (167.1 \ cc - 0.1 \ cc)$. Since $1 \ cc = 10^{-6} \ m^3$,$\Delta V \approx 167.1 	imes 10^{-6} \ m^3$. $\Delta W = 1.013 	imes 10^5 	imes 167.1 	imes 10^{-6} \approx 16.93 \ J$. $\Delta U = \Delta Q - \Delta W = 225.936 - 16.93 \approx 209 \ J$. Rounding to the nearest provided option,the change in internal energy is $208.7 \ J$.

$A$ 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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