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Question

Liquid A of specific heat capacity 0.84 J g−1 K−1 at a temperature of 50C is mixed with 200 g of a liquid B of specific heat capacity 2.1 J g−1 K−1 at 30C. The final temperature of the mixture becomes 42C. Find the mass of liquid A.


A

750 g

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B

333.33 g

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C

57.33 g

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D

400 g

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Solution

The correct option is A

750 g


Fall in temperature of liquid A = 5042C
Rise in temperature of liquid B = 4230C
Let 'm' g of liquid A be required.
Heat energy lost by m g of liquid A =m×0.84×(5042) J
Heat energy gained by 200 g of liquid B =200×2.1×(4230) J
Assuming there is no heat loss,
Heat energy lost by A = Heat energy gained by B
m×0.84×(5042)=200×2.1×(4230)
m=200×2.1×(4230)0.84×(5042)
m = 750 g


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