Answer to Question #166245 in General Chemistry for Quinn

Question

what temp change will occur when 1200 grams of water absorbs 13,650 calories of energy

Answer 1

Given,

Mass of water $m=1200g$

Energy $E=13650$ calories$=13650\times 4.2=57330J$

Specific heat of water $c=4.18J/gK$

Let $\Delta t$ be the change in temprature

As we know, $Q=mc\Delta t$

$\Rightarrow 57330=1200\times4.18\times \Delta t\\\Rightarrow\Delta t =\dfrac{57330}{5016}$

$\Rightarrow \Delta t=11.42K$