Question

$300$ grams of water at ${25}^{\circ }C$ is added to $100g$ of ice at $0^{\circ }C$. The final temperature of the mixture is ________.

Answer 1

The heat required for $100g$ of ice of $0^{\circ }C$ to change its temperature to $0^{\circ }C$
$=mL=100\times 80\times 4.2=33600J....\left(i\right)$
The heat released by $300g$ of water at ${25}^{\circ }C$ to change its temperature to $0^{\circ }C$
$=mc△=300\times 4.2\times 25=31,500J....\left(ii\right)$
Since the energy in Eq. (ii) is less than that of Eq. (i),

$\Delta H=33600-31,500=2100$

$mL=2100\Rightarrow m=80\times 4.2=2100$

$m=\frac{2100}{80\times 4.2}=6.25gm$

$\Rightarrow 393.75g\left(0^{0}Cwater\right)$

$\Rightarrow 6.25g\left(0^{0}ice\right)$

the final temperature will be $0^{\circ }C$.