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)

According to principle of calorimetry, heat given = heat taken.
Let heat given by water to cool upto $0^{\circ} C=m c \Delta \theta$
where $m$ is mass, $c$ is specific heat and $\Delta \theta$ 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
$m c \Delta \theta=m L+m c \Delta \theta'$
$100 \times 1 \times(50-\theta)=10 \times 80+10 \times 1 \times(\theta-0)$
$\Rightarrow 500-10 \theta=80+\theta$
$\Rightarrow 11 \theta=420$
$\Rightarrow \theta=\frac{420}{11}=38.2^{\circ} C$