Question

What mass of a liquid A of specific heat capacity 0.84 J K^-1 and at a temperature 40⁰C must be mixed with 100 g of a liquid B of specific heat capacity 2.1 J g^-1 K^-1 and at 20⁰C, so that final temperature of mixture becomes 32⁰C?

Answer 1

Let m be the mass of liquid A.

Assuming that there is no heat loss,

Heat energy given by A = Heat energy taken by B

or, m x 0.84 x ( 40 - 32) = 100 x 2.1 x ( 32 - 20)

or, m = `(100 xx 2.1 xx 12)/(0.84 xx 8)` = 375 g