Question

the volume of 1 kg of water at 100°C is 10^-3 m³ and the volume of 1 kg of steam at normal pressure is 1.671 m³ latent heat of steam is 2.3 *10^ 6 joule per kg and the normal pressure is 10^ 5 N/m² how much work will be done in converting 5 kg of water at hundred degree Celsius into steam at the same pressure and temperature

Answer 1

Answer on Question #82930, Physics / Molecular Physics | Thermodynamics

Question. The volume of $1\,kg$ of water at $100{}^{\circ}C$ is $10^{-3}\,m^{3}$ and the volume of $1\,kg$ of steam at normal pressure is $1.671\,m^{3}$ latent heat of steam is $2.3 \cdot 10^{6}$ joule per kg and the normal pressure is $10^{5}\,N/m^{2}$. How much work will be done in converting $5\,kg$ of water at hundred degree Celsius into steam at the same pressure and temperature.

Solution.

The work done is equal

$$W = \int dW = \int_{V_i}^{V_f} p\,dV = p(V_f - V_i) = 10^5 \cdot (5 \cdot 1.671 - 5 \cdot 10^{-3}) = 8.35 \cdot 10^5 J.$$

Answer. $W = 8.35 \cdot 10^5 J$.

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