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Question

Liquid A of specific heat capacity 1 J g−1 K−1 at a temperature of 50C is mixed with 100g of a liquid B of specific heat capacity 2 J g−1 K−1 at 30C.The final temperature of mixture becomes 40C. Find the mass of liquid A.


A

333g

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B

200 g

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C

33.3g

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D

400g

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Solution

The correct option is B

200 g


Fall in temperature of liquid A = 10C
Rise in temperature of liquid B = 10C
Let m g of liquid A be required.
Heat energy given by m g of liquid A = m×1×(10) J
Heat energy taken by 100g of liquid B = 100 ×2×(10) J
Assuming there is no heat loss,
Heat energy given by A = Heat energy taken by B

m×1×(10) = 100 ×2×(10)

m=100×2×(10)1×(10)
m= 200 g


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