Answer 2_5_12

Calculate how much energy is needed to melt, bring to the boil, and boil away 0.5 kg water.  

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

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

• DQ = mL = 0.5 kg x 334 000 J/kg = 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 x 4200 J kg^-1 K^-1 x 100 = 210000 J

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

• specific latent heat of vaporisation is rather higher, L_v = 2.3 x 10⁶ J/kg

• DQ = mL = 0.5 kg x 2.3 x 10⁶ J/kg = 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) 

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