A container has 2.4 litres of water at 20 C. heat required to boil water is
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volume of water in the container = 2.4Litres = 2.4 × 1000 ml = 2400 ml
we know, density of water = 1g/ml
so, mass of water in the container , m = 2400 × 1 = 2400g = 2.4 kg
we know, water boils at 0°C
so, final temperature = 100°C
initial temperature of water = 20°C
specific heat capacity of water , s = 4200 J/kg.°C
so, Heat required to boil water , H = ms∆T
= 2.4 kg × 4200J/kg.°C × (100 - 20)°C
= 2.4 × 4200 × 80
= 806400 J
= 8.064 × 10^5 J
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