Question

$10\, g$ of ice at $0^{\circ}\,C$ is mixed with $100\, g$ of water at $50^{\circ}\,C$. What is the resultant temperature of mixture?

• A. $31.2^{\circ}\,C$
• B. $32.8^{\circ}\,C$
• C. $36.7^{\circ}\,C$
• D. $38.2^{\circ}\,C$

Answer 1

Answer: (D)

Let heat given by water to cool upto
$0^{\circ} C =m c \Delta \theta$
where $m$ is mass, $c$ is specific heat, $\Delta \theta$ is temperature difference.
Heat taken by ice to melt $=m L$
where $L$ is latent heat
Also if $\theta$ is the temperature of the mixture then.
Heat taken = Heat given
$mc\Delta \theta = mL + mc \Delta \theta'$
$100 \times 1 \times (50 - \theta) = 10 \times 80 + 10 \times 1 \times (\theta - 0)$
$\Rightarrow 500 - 100 = 80 + 0$
$\Rightarrow 11 \theta = 420$
$ \Rightarrow \theta = \frac{420}{11} = 38.2^{\circ}$