# Question #2fa8b

$m$ gm of water of specific heat $s$ if suffers a temperature change of $\theta$ ,then heat($H$) required is $m \cdot s \cdot \theta$
Given, $m = 20 K g$ or $20000 g m$
$\theta = \left(95 - 15\right) = 80$
So, $H = 20000 \cdot 1 \cdot 80$ Calorie or $1600000$ Calorie