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Step-by-Step Solution
Step 1: Identify the Given Data
• Rate of water loss by evaporation = 1 g/min
• Water equivalent of pitcher = 0.5 kg
• Mass of water in the pitcher = 9.5 kg
• Initial temperature, $T_{\text{initial}} = 30^\circ C$
• Final temperature, $T_{\text{final}} = 28^\circ C$
• Latent heat of vaporization of water, $L = 580 \text{ cal g}^{-1}$
• Specific heat of water, $c = 1 \text{ kcal kg}^{-1} {}^\circ C^{-1}$
• $1 \text{ kcal} = 1000 \text{ cal}$
Step 2: Calculate the Total Mass Requiring Cooling
To cool both the water and the pitcher (accounted for here by its water equivalent), we sum their masses:
$ M_{\text{total}} = (9.5 + 0.5)\text{ kg} = 10.0\text{ kg} $
Step 3: Calculate the Heat to Be Removed
The temperature drop is $ (30^\circ C - 28^\circ C) = 2^\circ C $.
The heat to be removed, $ Q_1 $, is:
$ Q_1 = M_{\text{total}} \times c \times \Delta T $
Substitute the values:
$ Q_1 = 10.0\text{ kg} \times 1\text{ kcal kg}^{-1}{}^\circ C^{-1} \times 2^\circ C = 20\text{ kcal} $
Convert to calories:
$ Q_1 = 20\text{ kcal} \times 1000 \frac{\text{cal}}{\text{kcal}} = 2 \times 10^4 \text{ cal} $
Step 4: Heat Loss Due to Evaporation
If water evaporates at a rate of 1 g/min, in $ t $ minutes, $ t $ grams of water will evaporate.
The latent heat taken away by evaporation in $ t $ minutes is:
$ Q_2 = (t\text{ g}) \times 580 \text{ cal g}^{-1} = 580\, t \text{ cal} $
Step 5: Apply Heat Balance
To achieve the required cooling, the total heat removed by evaporation must equal the heat that needs to be extracted from the water plus pitcher. Thus:
$ Q_2 = Q_1 $
$ 580\, t = 2 \times 10^4 $
Step 6: Solve for the Time Required
Solving for $ t $:
$ t = \dfrac{2 \times 10^4}{580} = 34.48 \approx 34.5 \text{ minutes} $
Final Answer
The time required for the pitcher’s water to cool from $30^\circ C$ to $28^\circ C$ is about 34.5 minutes.
