How many grams of a liquid of specific heat 0.2 at a temperature `40^(@)C` must be mixedwith 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`

Options

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

Answer

D

Explanation

Heat loss = heat gain
`implies m_(l_1)s_(l_1)(40-T)=m_(l_2) s_(l_2)(T-20)`
`implies m_l xx 0.2 (40 -32)=100xx0.5(32-20)`
`implies m_l =(50xx12)/(0.2 xx 8)`=375 g.