Question

. How much heat must a refrigerator remove from 100 g of water at 25 º C to convert it to ice at 0º C? (Specific heat of water = 1 cal /gº C and heat of fusion of ice = 80 cal/g at 0º C)

Answer 1

Question#19615

. How much heat must a refrigerator remove from 100 g of water at 25 °C to convert it to ice at 0 °C? (Specific heat of water = 1 cal/g°C and heat of fusion of ice = 80 cal/g at 0 °C)

Solution:

Let:

$$m = 100\,g$$ $$T_2 = 0{}^\circ \text{C}$$ $$T_1 = 25{}^\circ \text{C}$$ $$c = 1\, \text{cal/g}{}^\circ\text{C}$$ $$\lambda = 80\, \text{cal/g}{}^\circ\text{C}$$

Q-?

$$Q = Q_1 + Q_2, \text{ were: } Q_1 - \text{heat of cooling water from } 25{}^\circ \text{C to } 0{}^\circ \text{C}, Q_2 - \text{heat of freezing}$$ $$Q = mc(T_1 - T_2) + \lambda m = m(c(T_1 - T_2) + \lambda)$$ $$Q = 100(1 * 25 + 80) = 10500\, \text{cal}$$

Answer: 10500 kal. Or 10.5 Kcal.