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)

$38.2^{\circ}\,C$

Answer 2

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 $mc\Delta \theta = mL + mc \Delta \theta'$ $100 \times 1 \times (50 - \theta) = 10 \times 80 + 10 \times 1 \times (\theta - 0)$ $\Rightarrow 500 - 100 = 80 + 0$ $\Rightarrow 11 \theta = 420$ $\Rightarrow \theta = \frac{420}{11} = 38.2^{\circ}$