Question

An earthen pitcher loses 1g of water per minute due to evaporation. If the water equivalent of pitcher is 0.5kg and the pitcher contains 9.5kg of water, calculate the time required for the water in the pitcher to cool to $28^{\circ} C$ from its original temperature of $30^{\circ}C$. Neglect radiation effects. Latent heat of vaporization of water in this range of temperature is $580 cal/g$ and specific heat of water is 1k cal/gC.
A. 38.6 min
B. 30.5 min
C. 34.5 min
D. 42.2 min

Answer 1

Specific heat can be defined as the heat required to raise the temperature of 1 gram mass of substance by one degree. It is a unique property of a substance which is used for thermodynamic calculations. As evaporation causes cooling, thus when water will come out and get evaporated, it will take away some heat from the system and hence the temperature will drop accordingly.

Complete answer:
First of all, we need to know the concept of water equivalent. Every container has its own specific heat because it is also made up of some material. Hence we have to consider the heat which goes to raise the temperature of the container. Practically, it’s impossible that all 100% heat will go to raise the temperature of water. Now, water equivalent may be defined as the mass of container whose temperature will rise by the same value on giving some heat as risen of water by the same heat. The definition seems confusing but the trick to avoid the confusion is to take the mass of water added with the mass of container when calculating heat using $Q=ms\Delta T$

Now, the question is so simple. What’s the heat needed to drop the temperature of the system by $(40^{\circ} - 38^{\circ} = 2^{\circ}C)$ is given by;
$Q = ms\Delta T$, where ‘m’ is the combined mass of water and container = 10kg
$\implies Q = 10 \dfrac{10^3}{10^{-3}} (2^{\circ} C) = 2\times 10^7\ cal$.
Now, the heated being ejected due to evaporation is given by $Q = mL$
$\implies \dfrac{dQ}{dt} = \dfrac{dm}{dt}L$
Given, $\dfrac{dm}{dt} = 1 g/min$
Thus, $\dfrac{dQ}{dt} = 1 \times 580 \times 10^3 cal$
Hence the heat ejected in time ‘t’ = $Q_{net} = 580 \times 10^3 t$
Now, as this much heat is needed to be ejected from the pitcher and it must be equal to $2\times 10^7\ cal$, so
$580\times10^3 t = 2\times 10^7$
$\implies t = \dfrac {2\times 10^7}{580\times 10^3} = 34.5 min$

So, the correct answer is “Option C”.

Note:
Keep special care of the units involved in the different quantities in the question. These types of large statement questions are meant to decrease the confidence of a student. But one must not panic and properly analyze the question and different concepts involved in it.