Properties Of Solids And Liquids Question 69

Question: 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 [Pb. PET 1999]

Options:

A) 175 gm

B) 300 g

C) 295 gm

D) 375 g

Show Answer

Answer:

Correct Answer: D

Solution:

Temperature of mixture $ \theta =\frac{m _{1}c _{1}{\theta _{1}}+m _{2}c _{2}{\theta _{2}}}{m _{1}c _{1}+m _{2}{\theta _{2}}} $

therefore $ 32=\frac{m _{1}\times 0.2\times 40+100\times 0.5\times 20}{m _{1}\times 0.2+100\times 0.5} $

therefore $ m _{1}=375gm $



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