Question

(iv)A container has 2.4 liters of water at 20 °C. The heat required to boil the water is
a. 1500 kJ
b. 1258 kJ
c. 1344 kJ
d. 698 kJ​

Answer 1

Answer:

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