The initial volume of the water is given as $3 \mathrm{m}^{3}$ and the mass is $0.1 \mathrm{kg}$. Therefore, the specific volume $v_{1}$ is given by $v_{1} = V_{1}/m = 3/0.1 = 30 \mathrm{m}^{3}/\mathrm{kg}$. At $40^{\circ} \mathrm{C}$, the saturation pressure
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