Question

40 gram of water at 60° C is poured into a vessel containing 50 g of water at 20° C . the final temperature of mixture is 30°C taking the specific heat capacity of water as 4.2 J g^-1 k^-1, calculate the thermal capacity of the vessel.

Answer 1

ANSWERS :-


Let thermal capacity of vessel be C' J K ^-1

Heat energy given by hot water

= mass of hot water * specific heat capacity * fall in temperature.

= 40 * 4.2 * ( 60 – 30 ) = 5040J .…...(i)

Heat energy taken by cold water

= mass of cold water * specific heat capacity * rise in temperature

= 50 * 4.2 * (30 - 20 ) = 2100J ..........(ii)

Heat energy taken by vessel

= thermal capacity of vessel * rise in temperature

= C ' * ( 30 - 20 ) = 10 C ' J

If there is no loss of heat energy,

Heat energy given by hot water

= Heat energy taken by cold water + Heat energy taken by vessel.

or,

5040 = 2100 + 10 C'
or,

10 C ' = 2940

or,

C ' = 2940/10 = 294 J k^-1

Thus,

thermal capacity of vessel = 294 J K ^-1

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❤BE BRAINLY ❤

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Answer 2

Explanation:

sorry for double writing