# To 10 liters of water at a temperature of 50 degrees, 6.9 MJ of heat was brought

**To 10 liters of water at a temperature of 50 degrees, 6.9 MJ of heat was brought, as a result of which part of the water evaporated. Determine the masses of the remaining water.**

V = 10 L = 0.01 m ^ 3.

ρ = 1000 kg / m ^ 3.

t1 = 50 ° C.

t2 = 100 ° C.

Q = 6.9 MJ = 6.9 * 10 ^ 6 J.

C = 4200 J / kg * ° C.

q = 2.3 * 10 ^ 6 J / kg.

m2 -?

The amount of heat Q will be spent on heating water from t1 to t2 and on the evaporation of some mass of water.

Q = Q1 + Q2.

Q1 = C * m * (t2 – t1).

We find the mass of water m by the formula: m = V * ρ.

Q1 = С * V * ρ * (t2 – t1).

Q2 = q * m1, where m1 is the mass of water that has evaporated.

Q = C * V * ρ * (t2 – t1) + q * m1.

m1 = (Q – C * V * ρ * (t2 – t1)) / q.

m1 = (6.9 * 10 ^ 6 J – 4200 J / kg * ° C * 0.01 m ^ 3 * 1000 kg / m ^ 3 * (100 ° C – 50 ° C)) / 2.3 * 10 ^ 6 J / kg = 2.08 kg.

m2 = V * ρ – m1.

m2 = 0.01 m ^ 3 * 1000 kg / m ^ 3 – 2.08 kg = 7.92 kg.

Answer: m2 = 7.92 kg of water left.