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