For the evaporation of water taken at C, 2000 kJ of heat was spent. How much water has evaporated?

For the evaporation of water taken at C, 2000 kJ of heat was spent. How much water has evaporated? The specific heat capacity of water is C = 4.2 kJ / kg * K, and the specific heat of vaporization is L = 2290 kJ / kg.

Q = 2000 kJ = 2000000 Joules – the energy spent on the evaporation of water;

T1 = 100 degrees Celsius – boiling point of water;

c = 4.2 kJ / kg = 4200 Joule / kilogram – specific heat of water;

q = 2290 kJ / kg = 2290000 Joule / kilogram is the specific heat of vaporization of water.

It is required to determine m (kilogram) – the mass of water.

Since the initial water temperature is not indicated in the problem statement, the problem will be solved first in general form, and then with the substitution of the initial temperature T0 equal to 20 degrees Celsius.

The total amount of spent energy will be equal to:

Q = Qheating + Qevaporation;

Q = c * m * (T1 – T0) + q * m;

Q = m * (c * (T1 – T0) + q);

m = Q / (c * (T1 – T0) + q).

Substituting the value of T0, equal to 20 degrees Celsius, we get:

m = 2,000,000 / (4200 * (100 – 20) + 2290000) = 2,000,000 / (336,000 + 2290000) = 2,000,000 / 2626,000 = 0.76 kilograms (rounded to one hundredth).

Answer: m = Q / (c * (T1 – T0) + q), at T0 = 20 degrees Celsius, the mass of water will be 0.76 kilograms.