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96.3 kJ of heat was generated when boiling water. How much water was evaporated?

User Molikh
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Here we have to get the amount of water evaporated which generate 96.3 kJ of heat at boiling condition.

The amount of water evaporated is 23.043 g.

We know H = m×s×t, where H = generated heat, m = mass of the material, s = specific heat of the material, t = temperature.

We know the specific heat of water is 4.179 J/g°C.

At boiling condition of water the temperature is 100°C.

The heat generated 96.3 kJ or 9630 J.

On plugging the values we get,

9630 J = m×4.179 J/g°C × 100°C

Thus, 9630 J = m × 417.9 J/g

m =
(9630 J)/(417.9 J/g)

m = 23.043 g.

User Alexej Magura
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