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

The molecules contain like 54 of water