Calculate the amount of heat required to convert 500 g of ice taken at a temperature of -10 ° C to steam.

m = 500 g = 0.5 kg.

t1 = – 10 ° С.

t2 = 0 ° C.

t3 = 100 ° C.

Cw = 4200 J / kg * ° С.

Cl = 2100 J / kg * ° С.

q = 3.4 * 10 ^ 5 J / kg.

r = 2.3 * 10 ^ 6 J / kg.

Q -?

The required amount of heat Q will be the sum: Q = Q1 + Q2 + Q3 + Q4, where Q1 is the amount of heat that is needed to heat the ice from t1 to the melting point t2, Q2 is the amount of heat that is needed to melt the ice, Q3 is the amount of heat , which is necessary to heat the resulting water from t2 to temperature t3, Q4 is the amount of heat that is needed to evaporate water at the boiling point.

Q1 = Cl * ml * (t2 – t1).

Q1 = 2100 J / kg * ° C * 0.5 kg * (0 ° C – (-10 ° C)) = 10500 J.

Q2 = q * ml.

Q2 = 3.4 * 10 ^ 5 J / kg * 0.5 kg = 170,000 J.

Q3 = Cw * ml * (t3 – t2).

Q3 = 4200 J / kg * ° С * 0.5 kg * (100 ° С – 0 ° С) = 210,000 J.

Q4 = r * ml.

Q4 = 2.3 * 10 ^ 6 J / kg * 0.5 kg = 1150000 J.

Q = 10500 J + 170,000 J + 210,000 J + 1150000 J = 1540500 J.

Answer: to convert ice into steam, you need Q = 1540500 J of thermal energy.