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Calculate the amount of heat (in kilojoules) required to vaporize 2.58 kg of water at its boiling point. Express the heat in
Suppose that 0.48 g of water at 25°C condenses on the surface of a 55-g block of aluminum that is initially at 25 °C. If the
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Answer #1


1) amount of heat required (q) = n*DHvap

n = no of mol of water = 2.58*10^3/18 = 143.33 mol

DHvap = enthalpy of vaporisation of water = 40.6 kj/mol

q = 143.33*40.6

   = 5819.2 kj

answer: 5.82*10^3 kj

2) heat lost by water = heat gained by Al

       m*s*DT+n*DHfus = m*s*DT

      0.48*4.18*(25-0)+(0.48/18)*6.01*10^3 = 55*0.903*(x-25)

    x = final temperature of Al block = 29 C

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