Question

What is the final temperature of a sample of water if 12.15J of energy is added to 475g of water at an initial temperature of 20.5°c?

Answer 1

Answer:

The final temperature of water is 20.5061 °C.

Explanation:

Let the final temperature of water be 'x'.

Given:

Heat added to water is, $Q_{in}=12.15\ J$

Initial temperature of water is, $T=20.5\ \°C$

Mass of water is, $m=475\ g$

Now, heat  is added to water and its temperature is increased. The temperature is increased because water absorbs all the heat.

Heat absorbed by water is given as:

$Q_{abs}=mc(x-T)$ where 'c' is specific heat capacity of water and its value is equal to 4.186 J/g °C.

Now, plug in the given values and simplify.

$Q_{abs}=475\times4.186\times(x-20.5)\\Q_{abs}=1988.35(x-20.5)$

Now, from law of conservation of energy, we know that:

Heat absorbed by water = Heat added to water

$Q_{abs}=Q_{in}\\1988.35(x-20.5)=12.15\\x-20.5=\frac{12.15}{1988.35}\\x-20.5=0.0061\\x=20.5+0.0061=20.5061\ \°C$

So, the final temperature of water is 20.5061 °C.