Question

Steam is passed through a copper vessel of 1000 g of specific heat 0.4 J/g-*C. 16.1 g of steam had condensed to at 10*C.Find the Specific latent heat of vapourization of steam

Answer 1

Question #61239, Physics / Molecular Physics | Thermodynamics

Steam is passed through a copper vessel of $1000\,\mathrm{g}$ of specific heat $0.4\,\mathrm{J/g}\cdot\mathrm{C}$. 16.1 g of steam had condensed to at $10{}^{\circ}\mathrm{C}$. Find the Specific latent heat of vapourization of steam

The answer to the question.

In accordance with the law of thermodynamics 1: the number of give off heat well received.

$$\begin{array}{l} Q_{1} = Q_{2}; \\ Q_{1} = C_{V} m_{1} \Delta T; \\ Q_{2} = \Delta H_{vap} m_{2}; \\ C_{V} m_{1} \Delta T = \Delta H_{vap} m_{2} \\ \Delta H_{vap} = \frac{C_{V} m_{1} \Delta T}{m_{2}} = \frac{0.4 \frac{J}{g} \cdot \mathrm{C} \cdot 1000\,\mathrm{g} \cdot (100{}^{\circ}\mathrm{C} - 10{}^{\circ}\mathrm{C})}{16.1\,\mathrm{g}} = 2236\,\mathrm{J}/\mathrm{g} \\ \end{array}$$

Answer: $\Delta H_{vap} = 2236\,\mathrm{J}/\mathrm{g}$

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