10 g of ice at - 10oC is mixed with 10 g of water at 0oC . The amount of heat required to raise the temperature of the mixture to 10oC is

To find the total heat required, we must account for three distinct stages. First, the 10 g of ice at -10°C must be raised to 0°C. Using the specific heat of ice (0.5 cal/g°C), this requires Q1 = 10 g × 0.5 cal/g°C × 10°C = 50 cal. Second, the 10 g of ice must melt into water at 0°C. Using the latent heat of fusion (80 cal/g), this requires Q2 = 10 g × 80 cal/g = 800 cal [1]. Third, the total mass of the mixture (now 20 g of water at 0°C) must be raised to 10°C. Using the specific heat of water (1 cal/g°C), this requires Q3 = 20 g × 1 cal/g°C × 10°C = 200 cal. The total heat is Q1 + Q2 + Q3 = 50 + 800 + 200 = 1050 cal.

Sources

1. [1] Physical Geography by PMF IAS, Manjunath Thamminidi, PMF IAS (1st ed.) > Chapter 22: Vertical Distribution of Temperature > Latent Heat > p. 294