Question

An 80-gram sample of water at 10C absorbs 400 calories of heat energy What is the final temperature of the water?

Answer 1

Answer : The final temperature of the water is $15^oC$.

Solution : Given,

Mass of water = 80 g

Initial temperature = $10^oC$

Heat energy = 400 calories = (400 × 4.18) J           (1 cal = 4.18 J)

Formula used :

$Q=m\times c\times \Delta T\\Q=m\times c\times (T_2-T_1)$

where,

Q = specific heat capacity

m = mass of water

c = heat capacity = $4.18J/^oC$

$T_1$ = initial temperature

$T_2$ = final temperature

Now put all the given values in this formula, we get

$400\times 4.18J=80g\times 4.18J/^oC\times (T_2-10^oC)\\T_2=15^oC$

Therefore, the final temperature of the water is $15^oC$.

Answer 2

We have that 80 grams of water absorbs 400 calories of heat energy. This energy will cause the temperature of the water to rise. The equation that relates the heat absorbed by  the water $Q$ to the change in temperature  $\Delta T$, the specific heat$c$ and the mass $m$ is ,

$Q=mc\Delta T.$

The known values in this problem are the temperature of the water $T=10C$, the mass of the water $m=80g$, the specific heat capacity of  water is $c=4.184J$. We also know that the water absorbs 400 calories of energy. We know that 1 calorie is equivalent to 4.184J of energy. From this we can gather that the water absorbs $400\times 4.184=1673.6J$ of energy.

We then have to make $\Delta T$, the subject of the formula and substitute in the given values to for the variables as shown below,

$Q=mc\Delta T\\=>\Delta T =\frac{Q}{mc}\\ \frac{1673.6J}{80g\times4.184j/gC}=5C$

The temperature of the water increases by 5C. The final temperature of the water is then 15C.