Question

How much heat energy is necessary to raise the temperature of 5 kg of water

from 20oC to 100oC ?

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Answer 1

Answer:

C is 4.186Jg°C.So,Q=750×4.186×85=266858J=266.858kJDT = temperature difference =100-20=80 0C. Answer is 266.858kj1 calorie = 4.18 Joules

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Answer 2

Explanation :-

↪Suppose, m= 5 kg and c= 1 kcal/ kg°C

↪Then, change in temperature Δ t = 100-20=80°C

↪Energy to be supplied to water = energy gained by water = mass of water × specific heat of water × change in temperature of water

↪${\sf{\red{Q = m × c × Δ t}}}$

↪Q = 5 × 1 × 80 = 5 × 80

↪${\sf{\blue{Q = 400 kcal}}}$

Therefore, the heat required to raise the temperature of water = 400 kcal.