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It depends a little on what temperature it is to start with.

The heat capacity of water is 4.184 J g–1 K–1 (which in calories is equal to 1 cal g–1 K–1), which means that it takes a little more than 4 Joules to heat 1 gram of water 1 °C. That means if the water is at room temperature (25 °C) you have to raise the temperature by 75 °C to reach the boiling point.
So 1 lb of water is is 453.59 grams, so that gives 1898.5 Joules (1.898 kJ) or 7943.3 calories (7.943 kcal) to heat 1 pound of water 75°C.

But that's just to get it to the boiling point. You then need more energy to convert the liquid water at 100 °C into gaseous water at 100 °C, and for that you need something called the heat of vaporization. For water that is 2258 J g-1(or 539.7 cal g-1).
So, multiplying by 453.59 g/lb, we get 1,024,357 J lb-1 (or 1024 kJ/lb), or 244.83 kcal per lb.

So most of the energy goes into turning water at 100 °C into steam at 100 °C, and so the answer is approximately 1025 kJ per pound, or 250 kcal per pound.


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18y ago

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