Question

Calculate the increase in the internal energy of 10 g of water when it is heated from $0^{\circ }C$ to ${100}^{\circ }C$ and converted into steam at 100 kPa. The density of steam $=0.6kg{m}^{-3}$. Specific heat capacity of water $=4200\text{J}k{g}^{-1}{}^{\circ }{C}^{-1}$ and the latent heat of vaporization of water $=2.25\times 106\text{J}k{g}^{-1}$.

Answer 1

Mass = 10 g = 0.01 kg.

$P={10}^5\text{kPa}$

$\Delta \text{Q}=Q{H}_{2}O{0}^{\circ }-{100}^{\circ }+Q{H}_{2}O-\text{steam}$

$=0.01\times 4200\times 100+0.01\times 2.5\times {10}^6$

$=4200+25000=29200$

$\Delta \text{W}=\text{P}\Delta \text{V}$

$\Delta \text{V}=\left(\frac{0.01}{0.6}\right)-\left(\frac{0.01}{1000}\right)=0.01699$

$\Delta \text{W}=\text{P}\Delta \text{V}=0.01699\times {10}^5=1699\text{J}$

$\Delta \text{Q}=\Delta \text{W}+\Delta \text{U}$

or $\Delta \text{U}=\Delta \text{Q}-\Delta \text{W}=29200-1699$

$=27501=2.75\times {10}^4\text{J}$