Question

A container has 2.4 liters of water at 20°C The heat required to boil the water is

Answer 1

Answer: 8.064 × 10⁵ J IS THE ANSWER.

Explanation:

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 × 10⁵ J.

HOPE IT HELPS..........................

THANK YOU.