Answer to Question 4

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First we need to calculate the energy needed to melt the ice:

  • specific latent heat of fusion, Lm = 334 000 J/kg.

  • DQ = mL = 0.5 kg 334 000 J kg-1 = 167 000 J

 

Next we need to find the energy supplied to bring the water to the boil:

  • Temp change = 100 oC - 0 oC = 100 oC (= 100 K)

  • DQ = mcDq = 0.5 kg 4200 J kg-1 K-1 100 K = 210000 J

 

Now we need to find out the energy needed to boil the water away:

  • specific latent heat of vaporisation is rather higher, Lv = 2.3 106 J kg-1

  • DQ = mL = 0.5 kg 2.3 106 J kg-1 = 1 150 000 J

 

Add them all together:

Total energy = 167 000 + 1 150 000  + 210000 = 1527000 J

 

How long would this take a 2 kW kettle?

Power = energy / time 

Time = energy / power = 1527000 J 2000 J/s = 763 s (about 13 minutes)