1.

How much heat energy is necessary to raise the temperature of 6 water from 30°C to 100°C​

Answer» ONG>Answer:

Given :

Mass =6kg

T1 =30∘C,T 2=100 ∘C

ΔT=100−30=70∘C

Q=m×C×ΔT

where C= specific heat capacity of water

=4200J/(KGK)

Q=6×4200×70

=1764000 Joule.

=1764KJ



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