How many degrees will the temperature of water, the mass of which is 20 kg, change, if 80% of all the energy released during the combustion of natural gas with a mass of 15 g is transferred to it?
m = 20 kilograms is the mass of water;
c = 4200 J / (kg * C) – specific heat capacity of water;
m1 = 15 grams = 0.015 kilograms is the mass of natural gas;
q = 46.1 * 10 ^ 6 J / kg – specific heat of combustion of natural gas;
n = 80% = 0.8 is the percentage of energy from the combustion of natural gas that is transferred to the water.
It is required to determine dt (degree Celsius) – how much the water temperature will increase.
When natural gas is burned, energy will be released equal to:
Q = q * m1.
According to the condition of the problem, energy will be transferred to heating the water:
Q1 = n * Q = n * q * m1.
Then the change in water temperature will be equal to:
dt = Q1 / (c * m) = n * q * m1 / (c * m) = 0.8 * 46.1 * 10 ^ 6 * 0.015 / (4200 * 20) = 553200/84000 = 6.6 ° Celsius.
Answer: The water temperature will increase by 6.6 ° Celsius.
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