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

Options

A. `175 gm`
B. `300 g`
C. `295 gm`
D. `375 g`

Answer

D

Explanation

Temperature of mixture `theta = (m_(1)c_(1)theta_(1)+m_(2)c_(2)theta_(2))/(m_(1)c_(1)+m_(2)theta_(2))`
`rArr 32 = (m_(1)xx 0.2 xx 40 + 100 xx 0.5 xx20)/(m_(1)xx0.2+100xx0.5)`
`rArr m_(1) =375 gm`.