Answer to Question #111539 in Physical Chemistry for Sharad Sailesh

Question

15. A pot contains 15g of water at 25°C. If this water absorbs 489.5J of energy, what will its final temperature be? Will it be boiling?

Answer 1

The quantity of heat needed to raise the temperature of the water to the boiling point is:

$Q = cm∆T = cm(100 \text{°C} - 25\text{°C})$ ,

where $c$ is the specific heat capacity of water. If we neglect the dependence of heat capacity on temperature, we can take the value of  4179.6 J/(kg K). Note that as we use a difference temperature $∆T$, it has the same numerical value in celsius and in kelvins. Therefore, the energy needed is:

$Q = 4179.6 (\text{J/(kg K)})·15·10^{-3}(\text{kg})·(75 \text{K}) = 4702 (\text{J})$

As we can see, the quantity of heat needed to raise the temperature of the water to the boiling point is much higher than the quantity of heat absorbed: 4702 J>>489.5 J.

Therefore, the water will not be boiling. Its final temperature will be:

$T_2 = \frac{Q}{cm} + T_1 = \frac{489.5 \text{J}}{ 4179.6 (\text{J/(kg K)})·15·10^{-3} (\text{kg})} +25 \text{°C} = 32.8 \text{°C}$

Answer: No, the water will not be boiling. Its final temperature will be 32.8 °C.