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: Concept
The heat lost by the total water mass system through evaporation equals the heat required to lower the temperature of the water along with the pitcher itself: $m_{\text{evap}} \cdot L = (m_{\text{water}} + W_e) \cdot s \cdot \Delta \theta$, where $W_e$ is the water equivalent of the pitcher and $s$ is the specific heat of water.

Step 2: Meaning
Evaporation carries away thermal energy because changing liquid to vapor requires the latent heat of vaporization ($L = 500 \text{ cal g}^{-1}$). The rate of evaporation given is $R = 1 \text{ g min}^{-1}$.

Step 3: Analysis
Total effective mass of the water system is $M = 9.5 \text{ kg} + 0.5 \text{ kg} = 10 \text{ kg} = 10000 \text{ g}$. The required drop in temperature is $\Delta \theta = 30^\circ\text{C} - 28^\circ\text{C} = 2^\circ\text{C}$. Given specific heat of water $s = 1 \text{ cal g}^{-1}\text{}^\circ\text{C}^{-1}$, the total heat to be removed is $Q = M \cdot s \cdot \Delta \theta = 10000 \times 1 \times 2 = 20000 \text{ calories}$. Let $t$ be the time in minutes. The total mass evaporated in time $t$ is $m_{\text{evap}} = 1 \times t = t \text{ grams}$. Heat lost by evaporation is $Q_{\text{lost}} = t \times 500$. Equating the two quantities: $500t = 20000 \implies t = \frac{20000}{500} = 40 \text{ minutes}$.

Step 4: Conclusion
Therefore, the total time required for the cooling process inside the pitcher is 40 minutes.

Final Answer: (C)