Question

A container holds 2.0 of liquid water. The water absorbs 477 kJ of energy. If the water's initial temperature was 298 K,what is it's final temperature?

Answer 1

Answer:

$T_{final}=82C$

Explanation:

Here mass of water not given directly but volume of water is given and we already know the density of water so we will use the volume density relationship to find mass.

volume=2L

density=1kg/L

$density=\frac{mass}{volume}$

$mass=density\times volume$

mass=2kg=2000gram

initial temperature=298K=25C

specific heat of water (C)=4.186 joule/gram

heat energy lose or gain=$mC\Delta T$

$477\times 10^3 J=2000g\times 4.186J/g \times \Delta T$

$\Delta T=57 C$

$T_{final}-T{initial}=57$

$T_{final}=82C$