How much heat is required to convert 74 grams of ice at - 4 C into 74 grams of water at 52 C

Hf=334 j/g

To calculate the heat required to convert 74 g of ice at - 4 °C to 52 °C, we need to know the specific heat capacity of ice and water. We can use the following formula:

q = mcΔT

m = mass of substance

c = specific heat capacity

ΔT = change in temperature

The specific heat capacity of ice is 2.108 J/g°C and for water is 4.187 J/g°C. We will first calculate the heat to convert ice from - 4 °C to 0 °C. Then we will calculate the heat to increase the temperature of water from 0 °C to 52 °C.

From - 4 °C to 0 °C:

q = (74 g) (2.108 J/g°C) (4°C)

q = 624 J

From 0 °C to 52 °C:

q = (74 g) (4.187 J/g°C) (52°C)

q = 16,112 J

The total heat required to convert 74 g of ice at - 4 °C to 74 g of water at 52 °C is the combined calculated heats:

Total heat = 16,112 + 624

Total heat = 16,736 J = 16.7 kJ