Question

In an energy recycling process x gram of steam at 100 ◦ C becomes water at 100 ◦ C which converts Y gram of ice at 0◦ C into water at 100 ◦ C. The ratio of X and Y will be

Answer 1

Answer on Question #41180 – Physics – Molecular Physics | Thermodynamics

In an energy recycling process X gram of steam at $100{}^{\circ}\mathrm{C}$ becomes water at $100{}^{\circ}\mathrm{C}$ which converts Y gram of ice at $0{}^{\circ}\mathrm{C}$ into water at $100{}^{\circ}\mathrm{C}$. The ratio of X and Y will be

Solution:

$\lambda = 334\frac{\mathrm{kJ}}{\mathrm{g}} - \text{heat of fusion for the water};$

$q = 2260\frac{\mathrm{kJ}}{\mathrm{g}} - \text{heat of evaporation for the water};$

$c = 4180\frac{\mathrm{J}}{\mathrm{g}\cdot{}^{\circ}\mathrm{C}} - \text{heat capacity of water};$

$\Delta T = 100{}^{\circ}\mathrm{C} - \text{change in temperature};$

$$Q_{\text{lost}} = Q_{\text{gain}} \quad (1)$$

The amount of heat lost by steam:

$$Q_{\text{lost}} = q \cdot X \quad (2)$$

The amount of heat gained by ice:

$$Q_{\text{gain}} = Y \cdot \lambda + cY\Delta T \quad (3)$$

(3) and (2) in (1)

$$q \cdot X = Y \cdot \lambda + cY\Delta T$$ $$q \cdot X = Y (\lambda + c\Delta T)$$ $$\frac{Y}{X} = \frac{\lambda + c\Delta T}{q} = \frac{2260\frac{\mathrm{kJ}}{\mathrm{g}}}{334\frac{\mathrm{kJ}}{\mathrm{g}} + 4180\frac{\mathrm{J}}{\mathrm{g}\cdot{}^{\circ}\mathrm{C}} \cdot 100{}^{\circ}\mathrm{C}} = 3$$ $$Y = 3X$$

Answer: The ratio of X and Y will be: $Y = 3X$.

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