# Calculate the amount of heat required to convert 20 kg of ice at -4 degrees Celsius to water at 100 degrees Celsius.

calculate the amount of heat required to convert 20 kg of ice at -4 degrees Celsius to water at 100 degrees Celsius

m = 20 kg.

t1 = – 4 ° С.

t2 = 0 ° C.

t3 = 100 ° C.

Cw = 4200 J / kg * ° С.

Cl = 2100 J / kg * ° С.

q = 3.4 * 10 ^ 5 J / kg.

Q -?

The required amount of heat Q will be the sum: Q = Q1 + Q2 + Q3, where Q1 is the amount of heat that is needed to heat ice from t1 to the melting point t2, Q2 is the amount of heat that is needed to melt ice, Q3 is the amount of heat that it is necessary to heat the obtained water from t2 to temperature t3.

Q1 = Cl * ml * (t2 – t1).

Q1 = 2100 J / kg * ° C * 20 kg * (0 ° C – (- 4 ° C)) = 168000 J.

Q2 = q * ml.

Q2 = 3.4 * 10 ^ 5 J / kg * 20 kg = 6800000 J.

Q3 = Cw * ml * (t3 – t2).

Q3 = 4200 J / kg * ° C * 20 kg * (100 ° C – 0 ° C) = 8400000 J.

Q = 168000 J + 6800000 J + 8400000 J = 15368000 J.

Answer: to melt ice and heat water, you need Q = 15368000 J of thermal energy.