Question

When $100g$ of liquid $A$ at $100^\circ C$ is added to $50g$ of liquid $B$ at temperature $75^\circ C$ , the final temperature of the mixture becomes $90^\circ C$ . Similarly, if $100g$ of liquid $A$ at $100^\circ C$ is added to $50g$ of liquid $B$ at $50^\circ C$ , what will be the final temperature in $^\circ C$ ?

Answer 1

In first case according to principle of calorimetry,
heat lost by liquid $A=$ heat gained by liquid $B$
$\therefore m_{A}c_{A}ΔT_{A}=m_{B}c_{B}ΔT_{B}$
$\therefore 100\times c_{A}\left(\right.100-90\left.\right)=50\times c_{B}\left(\right.90-75\left.\right)$
$\therefore 1000c_{A}=50\times 15c_{B}$
$\therefore 4c_{A}=3c_{B}\ldots .\left(\right.i\left.\right)$
Similarly, in second case,
$100\times c_{A}\left(\right.100-T\left.\right)=50\times c_{B}\left(\right.T-50\left.\right)$
$\therefore 4c_{A}\left(\right.100-T\left.\right)=2c_{B}\left(\right.T-50\left.\right)$
Using equation (i),
$3c_{B}\left(\right.100-T\left.\right)=2c_{B}\left(\right.T-50\left.\right)$
$\therefore 300-3T=2T-100$
$\therefore 5T=400$
$\therefore T=80^\circ C$