The steam heating radiator condensed 10 kg of water vapor at 100 degrees, and the water came out of the radiator

The steam heating radiator condensed 10 kg of water vapor at 100 degrees, and the water came out of the radiator at 80 degrees. How much heat has the radiator transferred to the ambient air?

Given:

m = 10 kilograms is the mass of water that has been condensed from water vapor in the radiator;

t = 100 ° degrees Celsius – water vapor temperature;

t1 = 80 ° degrees Celsius – the temperature of the water leaving the radiator;

c = 4200 Joule / (kg * C) – specific heat of water;

q = 2.3 MJ / kg = 2300000 Joule / kilogram is the specific heat of vaporization of water.

It is required to determine Q (Joule) – the amount of heat that the radiator transferred to the environment.

The amount of heat released as a result of steam condensation will be equal to:

Q 1 = q * m = 2,300,000 * 10 = 23,000,000 Joules.

The amount of heat released as a result of water cooling will be equal to:

Q2 = c * m * (t – t1) = 4200 * 10 * (100 – 80) = 4200 * 10 * 20 = 42000 * 20 = 840000 Joules.

Total heat released:

Q = Q1 + Q2 = 23,000,000 + 840,000 = 23,840,000 Joules = 23.84 MJ.

Answer: the radiator transferred an amount of heat equal to 23.84 MJ to the environment.