Question

540 calories of heat is required to vaporize 1 gm of water at 100C.determine the entropy change involved in vaporizing 5 gm of water?

Answer 1

540 calories of heat is required to vaporize 1 gm of water at 100°C. Determine the entropy change involved in vaporizing 5 gm of water?

Solution.

$$L = 540 \frac{cal}{g}, \quad t = 100{}^\circ \text{C}, \quad m = 5g = 0.005kg;$$ $$\Delta S - ?$$

$L = 540 \frac{cal}{g}$ - the latent heat of vaporize of water.

Converting the latent heat to joules per grams:

$$1 \text{ cal} = 4.1868J;$$ $$L = 540 \frac{cal}{g} (4.1868J) = 2260.872 \frac{J}{g}.$$

Converting the latent heat to joules per kilograms:

$$L = 2260.872 \frac{J}{g} \left(\frac{1000g}{1kg}\right) = 2260872 \frac{J}{kg}.$$ $$L = 2260872 \frac{J}{kg}.$$

The entropy change is:

$$\Delta S = \frac{\Delta Q}{T}.$$

$\Delta Q$ - the heat that is required to vaporize 5 g of water.

$$\Delta Q = mL.$$

Thermodynamic temperature:

$$T = (t + 273{}^\circ \text{C})K;$$ $$t = 100{}^\circ \text{C};$$ $$T = (100{}^\circ \text{C} + 273{}^\circ \text{C})K;$$ $$T = 373K.$$ $$\Delta S = \frac{mL}{T}.$$ $$\Delta S = \frac{0.005kg \cdot 2260872 \frac{J}{kg}}{373K} = 30.3 \frac{J}{K}.$$

Answer: The entropy change is $\Delta S = 30.3 \frac{J}{K}$.