Question

Indian style of cooling drinking water is to keep it in a pitcher having porous walls. Water comes to the outer surface very slowly and evaporates. Most of the energy needed is taken from the water itself and the water is cooled down. Assume that a pitcher contains $10kg$ of water and $0.2g$ of water comes out per second. Assuming no backward heat transfer from the atmosphere to the water, calculate the time in which the temperature decreases by ${{5}^{0}}C$. Specific heat capacity of water $=4200J.k{{g}^{-1}}{{K}^{-1}}$ and latent heat of vaporization of water $=2.27\times {{10}^{6}}J.k{{g}^{-1}}$

Answer 1

Hint: This problem can be solved by calculating the total amount of heat energy that must be removed from $10kg$ of water to make its temperature drop by ${{5}^{0}}C$ and then dividing it by the energy absorbed by the water for evaporation per second. Thus, we will get the total time in seconds for the required process.

Formula used:
Amount of heat $\left( H \right)$ that has to be removed to decrease the temperature of water of mass $m.kg$ by ${{T}^{0}}C$ is given by
$H=mST$
where $S$ is the specific heat capacity of water equal to $=4200J.k{{g}^{-1}}{{K}^{-1}}$.
The amount of heat ${{H}_{evap}}$ required by $m.kg$ of water to evaporate is given by
${{H}_{evap}}=mL$
where $L$is the latent heat of vaporization of water equal to $2.27\times {{10}^{6}}J.k{{g}^{-1}}$

Complete step by step answer:
In a water pitcher, water trickles to the outside and uses the heat from the rest of the water inside the pitcher, to evaporate, thereby, cooling the water inside the pitcher.
We can solve this problem by calculating the total heat that has to be removed from the water in the pitcher to make its temperature drop by ${{5}^{0}}C$and then equating it to the heat absorbed by the water that is getting evaporated per second and the total time taken for it, thereby giving us the required time.

Amount of heat $\left( H \right)$ that has to be removed to decrease the temperature of water of mass $m.kg$ by ${{T}^{0}}C$ is given by
$H=mST$ --(1)
where $S$is the specific heat capacity of water equal to $=4200J.k{{g}^{-1}}{{K}^{-1}}$.
The amount of heat ${{H}_{evap}}$ required by $m.kg$of water to evaporate is given by
${{H}_{evap}}=mL$ --(2)
where $L$is the latent heat of vaporization of water equal to $2.27\times {{10}^{6}}J.k{{g}^{-1}}$

Therefore, let us analyze the question.
Total mass of water inside the pitcher$M=10kg$.
Required temperature drop $\Delta T={{5}^{0}}C$.
Specific heat capacity of water $S=4200J.k{{g}^{-1}}{{K}^{-1}}$.
Hence, using (1), the total heat $H$ that has to be removed will be
$H=10\times 4200\times 5$ ($\because K=273+C$ $\therefore \Delta K=\Delta C$)
$\therefore H=210000=2.1\times {{10}^{5}}J$ --(3)

Now, per second $0.2g$ of water is getting evaporated.

Mass of water getting evaporated per second $m=0.2g=2\times {{10}^{-4}}Kg$ $\left( \because 1g={{10}^{-3}}Kg \right)$
Latent heat of vaporization of water $L=2.27\times {{10}^{6}}J.k{{g}^{-1}}$
Therefore, heat absorbed per second ${{H}_{evap}}$ for the evaporation from the water inside the pitcher, using (2) will be,
${{H}_{evap}}=2\times {{10}^{-4}}\times 2.27\times {{10}^{6}}=454J/s$
Hence, let the required amount of time to drop the temperature of the water inside the pitcher by ${{5}^{0}}C$ be $t$ seconds. Hence, the total heat absorbed during evaporation in this time will be
${{H}_{evap}}\times t$
$=454\times t\text{ }J$ --(4)
Now, since there is no backward heat transfer from the atmosphere to the water, the water absorbs heat from the rest of the water inside the pitcher to evaporate.
Hence, $H={{H}_{evap}}\times t$
Therefore, using (4) and (5),
$\therefore 2.1\times {{10}^{5}}=454\times t$
$\therefore t=\dfrac{2.1\times {{10}^{5}}}{454}=462.56s$
$\therefore t=\dfrac{462.56}{60}=7.71\text{ minutes}$ $\left( \because 1\text{ minute = 60 seconds} \right)$
Hence, the total required time is $7.71\text{ minutes}$.

Note: Students might be a bit perplexed and think that if the water is evaporating this means that the total mass of the water is decreasing and hence the heat required for bringing the temperature down should also decrease. Though this is not wrong, however for ease of understanding the problem and also since the rate of trickling out of the water is negligible in comparison to the total mass of the water inside the pitcher, this is neglected.
Most typical calorimetry problems involve heat transfer between bodies. To solve such problems, it is easier that the student separately calculates the amounts of heat absorbed and released by the different bodies, if any and then equates or finds a ratio between the quantities according to the requirements of the problem.