Question

please help me with this

Answer 1

Heat required=m.c.(∆T)




water starts boiling at 100℃

so ∆T=100-20=80℃

mass of water is same as it's volume.

mass=2.5 Kg.

C=4.186 joules/g℃

$heat = 2.5 \times {10}^{3} \times 4.186 \times 80 \\ 837.2kj$
➡️Approximately equal to 840Kj.

Hope it will help you✔️