Q6. If the temperature of 5 Kg of water is
increased by 20°C, the quantity of heat
required is:
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Answer:
Given :
Mass =5kg
T1=20∘C,T2=100∘C
ΔT=100−20=80∘C
Q=m×C×ΔT
where C= specific heat capacity of water
=4200G/(kgK)
Q=5×4200×80
=1680000 Joule.
=1680KG
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