Question

$10g$ of ice at $0^{\circ }C$ is mixed with $100g$ 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)

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