From water taken at 10 degrees, you need to get 15 kg of water vapor at 100 degrees.

From water taken at 10 degrees, you need to get 15 kg of water vapor at 100 degrees. How much bituminous coal must be burned for this if the heater efficiency is 20 percent?

Given:

m = 15 kilograms is the mass of water;

T0 = ​​10 degrees Celsius – initial water temperature;

T1 = 100 degrees Celsius – boiling point of water;

c = 4300 J / (kg * C) – specific heat capacity of water;

q1 = 2.3 * 10 ^ 6 J / kg is the specific heat of vaporization of water;

q2 = 29 * 10 ^ 6 J / kg – specific heat of combustion of coal;

n = 20% = 0.2 – heater efficiency.

It is required to determine the amount of coal m1 (kilogram).

To heat and evaporate water, you need to expend energy:

Q = Q1 + Q1 = c * m * (T1 – T) + q1 * m =

= 4200 * 15 * 90 + 2.3 * 10 ^ 6 * 15 = 5.67 * 106 + 34.5 * 10 ^ 6 = 40.17 * 106 Joules.

Taking into account the efficiency of the heater, it is only necessary to obtain energy from coal combustion:

Q total = Q / n = 40.17 * 10 ^ 6 / 0.2 = 200.85 * 10 ^ 6 Joules.

Then the mass of coal will be equal to:

m = Q total / q2 = 200.85 * 10 ^ 6/29 * 10 ^ 6 = 6.9 kilograms.

Answer: you need to burn 6.9 kilograms of coal.