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 $
