Question

A conducting tube is passing through a bath. A liquid at temperature ${90}^{\circ }C$ and specific heat s is entering at one end of tube. The rate of flow of liquid is 1kg/s and exit temperature is ${50}^{\circ }C$. In bath another liquid having specific heat 2s and inlet temperature ${20}^{\circ }C$ is entering at a rate of 2kg/s. If the exit temperature of liquid coming out of the bath is 10x. Find x (Assume steady-state condition) ___

Answer 1

Heat lost by liquid flowing through the tube in 1 second = Heat gained by liquid in bath in 1 second
$\Rightarrow 1\times s\times (90-50)=2\times 2s\times (10x-20)$

$40=4(10x-20)$

$80+40=40x$

$120=40x$

$x=3$