Question

An earthen pitcher containing 9.5 kg of water loses one gram of water per minute due to evaporation. If water equivalent of pitcher is 0.5 kg. The time required to cool the water in pitcher from $30^{\circ}\text{C}$ to $28^{\circ}\text{C}$ is (Neglect radiation effect and Take latent heat of vaporization 500 cal $\text{g}^{-1}$)

• A. 30 min
• B. 60 min
• C. 40 min
• D. 20 min

Hint:

Always combine the actual mass of the liquid with the water equivalent ($W_e$) of its vessel to find the total thermal mass.

Answer 1

The Correct Option is C

Step 1: Balance the heat.
The heat carried away by evaporation cools the water and pot. So $m_{ev}L = (m_{w} + W_{e})\,s\,\Delta\theta$.

Step 2: Why evaporation cools.
Turning liquid into vapour needs latent heat $L = 500$ cal g$^{-1}$. This heat is taken from the water, so it cools down.

Step 3: Find the total effective mass.
Water $9.5$ kg plus water equivalent $0.5$ kg gives $10$ kg $= 10000$ g.

Step 4: Heat that must leave.
Temperature falls by $\Delta\theta = 2^{\circ}$C, and $s = 1$ cal g$^{-1}\,^{\circ}$C$^{-1}$. \[ Q = 10000\times 1\times 2 = 20000 \text{ cal} \]
Step 5: Heat removed by evaporation in time $t$.
At $1$ g per minute, in $t$ minutes mass $= t$ g, so heat removed $= 500t$ cal.

Step 6: Set them equal and solve.
\[ 500t = 20000 \quad\Rightarrow\quad t = 40 \text{ min} \]So it takes $40$ minutes, which is option 3.
\[ \boxed{t = 40 \text{ min}} \]