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

Explanation

The liquid at temperature `40^@C` will give heat to the other liquid at temperature `20^@C`. The temperature of the mixture becomes `32^@C`. Now
heat taken=`100xx0.5xx(32-20)=60` cal
heat given = `mxx0.2xx(40-32)=1.6 m` cal
Now, heat taken= heat given
600 cal = 1.6 m or, m=600/1.6 = 375 gm