What mass of a liquid A of specific heat capacity 0.84 J `g^(-1) K^(-1)` at a temperature `40^@`C must be mixed with 100 g of a liquid B of specific heat capacity 2.1 J `g^(-1) K^(-1)` at `20^@`C, so that the final temperature of mixture becomes `32^@`C ?

Explanation

Let m g of liquid A be required.
Fall in temperature of liquid A = `(40 - 32)^@`C
Rise in temperature of liquid B = `(32 - 20)^@` C
Heat energy given by m g of liquid A = m `xx` 0.84 `xx` (40 - 32) J (i)
Heat energy taken by 100 g of liquid B = 100 `xx` 2.1 `xx` (32 – 20) J .(ii)
Assuming that there is no heat loss,
heat energy given by A = heat energy taken by B
or m `xx` 0.84 `xx` 8 = 100 `xx` 2.1 `xx` 12
`therefore m= (100xx2.1xx12)/(0.84xx8)` = 375 g