Explanation

(c) Let $m$ gram of water, whose temperature is $\theta$(> 30$℃),\ $ be added to 20 g of water at 30$℃$. If $m\ \times 1\left( \theta - \theta_{0} \right) = 20 \times 1(\theta_{0} - 30)$
($m$+$20)\theta_{0} = 60 + m\theta$
$\theta_{0} = \ \frac{600 + m\theta}{20 + m}$
For $\theta_{0}\ to$ be maximum $m$ should be small and $\theta$ should be large