How much heat is needed to turn 800 g of water into steam at the boiling point?

Let Q is the required amount of heat, L is the specific heat of vaporization of water, m is the mass of water. We know that Q = L * m, since water is taken at boiling point, which means that it is not required to heat it, we substitute numerical values (we know the mass, the specific heat of vaporization of water is a tabular value), and we get that Q = 0.8 kg * 2260 kJ / kg = 1808 kJ. This means that the required amount of heat is 1808 kJ.
Answer: 1808 kJ of heat.

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