m1 = 10 kilograms is the mass of water;
m2 = 1 kilogram – steam mass;
c = 4200 J / (kg * C) – specific heat capacity of water;
q = 2260 * 10 ^ 3 J / kg – specific heat of vaporization (condensation) of water.
It is required to determine dt (degree Celsius) – how many degrees the water will be heated due to the energy obtained during the condensation of water vapor.
During condensation of water vapor, energy will be released equal to:
E = q * m2 = 2260 * 10 ^ 3 * 1 = 2260000 Joules.
Then, the temperature change will be equal to:
dt = E / (c * m1) = 2260000 / (4200 * 10) = 2260000/42000 = 2260/42 = 53.8 ° C (the result has been rounded to one decimal place).
Answer: The water will heat up to 53.8 ° Celsius.
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