- Home
- Standard 11
- Physics
An aluminium piece of mass $50 \,g$ initially at $300^{\circ} C$ is dipped quickly and taken out of $1 \,kg$ of water, initially at $30^{\circ} C$. If the temperature of the aluminium piece immediately after being taken out of the water is found to be $160^{\circ} C$, the temperature of the water ............ $^{\circ} C$ Then, specific heat capacities of aluminium and water are $900 \,Jkg ^{-1} K ^{-1}$ and $4200 \,Jkg ^{-1} K ^{-1}$, respectively.
$165$
$45$
$31.5$
$28.5$
Solution
(c)
As heat lost by aluminium piece $=$ heat gained by water
$\left\{m s\left(T_i-T_f\right)\right\}_{\text {aluminium }}=\left\{m s\left(T_f-T_i\right)\right\}_{\text {water }}$
Substituting given values, we get
$\Rightarrow 50 \times 10^{-3} \times 900 \times(300-160)$ $=1 \times 4200 \times(T-30)$
$\Rightarrow 6300=4200(T-30)$
$\Rightarrow T=30+1.5$
$\text { or } T=31.5^{\circ} C$
So, temperature of water after taking out aluminium piece is $31.5^{\circ} C$.
