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Q.

How many grams of a liquid of specific heat 0.2 at a temperature 40°C must be mixed with 100 gm of a liquid of specific heat of 0.5 at a temperature 20°C, so that the final temperature of the mixture becomes 32°C

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a

300 g

b

375 g

c

295 gm

d

175 gm

answer is D.

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Detailed Solution

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Temperature of mixture θ = m1c1θ1+m2c2θ2m1c1+m2 c2

⇒32 = m1×0.2×40+100×0.5×20m1×0.2+100×0.5

⇒m1 = 375 gm

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