Question

If 20 grams of water at 20°C is supplied 20 Cal of heat energy then temperature of water is ________​

Answer 1

Answer:

Heat absorbed by a material = m*Cp*(T2-T1)

m - mass of the substance (kg)

Cp - Specific heat of the substance (J/kg.C)

T1 - Initial temperature

T2 - Final temperature

Now as your said, 20 cal are added to the water, which is equivalent to 83.68J.

Equalate the heat absorbed to the formula..

83.68 = 0.02*4183*(T2-20)

Solving this we can seebthere is unit rise in temperature. T2 = 21C

Thus we can say specific heat of water is 1cal/g.C

And that's verified.

Thank you.

Explanation:

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