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How much heat (in kJ) is required to convert 431 g of liquid H2O at 23.6°C...

How much heat (in kJ) is required to convert 431 g of liquid H2O at 23.6°C into steam at 148°C? (Assume that the specific heat of liquid water is 4.184 J/g·°C, the specific heat of steam is 2.078 J/g·°C, and that both values are constant over the given temperature ranges. The normal boiling point of H2O is 100.0°C. The heat of vaporization (ΔHvap) is 40.65 kJ/mol.)

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Answer #1

no of moles of H2O   = W/G.M.Wt

                                   = 431/18   = 23.94moles

The heat energy is required to water convert 23.60C to 1000C

q1 = mct

       = 431*4.184*(100-23.6)

        = 137772J

         = 137.772KJ

The heat of vaporization

q2   = n Hvap

         = 23.94*40.65

           = 973.16KJ

The heat energy is required water convert to 1000C to 1480C

q3   = mct

       = 431*2.078*(148-100)

         = 42990J

         = 42.99KJ

Total heat energy is required

q   = q1 + q2 + q3

       = 137.772 + 973.16+42.99

      = 1154KJ

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