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How much heat energy is necessary to raise the temperature of 6 water from 30°C to 100°C |
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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 Q=6×4200×70 =1764000 Joule. =1764KJ |
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