To determine the time required for the water in the earthen pitcher to cool from 30°C to 28°C, we need to consider the heat loss due to evaporation and the heat capacity of the water. Let's break this down step by step.
Understanding the Problem
We know the following details:
- Water loss due to evaporation: 1 g/min
- Water equivalent of the pitcher: 0.5 kg (or 500 g)
- Initial water mass: 9.5 g
- Latent heat of vaporization (Lv): 580 cal/g
- Specific heat of water: 1 cal/g°C
- Initial temperature: 30°C
- Final temperature: 28°C
Calculating the Heat Loss
To find the time required for the temperature drop, we first need to calculate the total heat loss as the water cools from 30°C to 28°C. The heat lost by the water can be calculated using the formula:
Q = m × c × ΔT
Where:
- Q = heat lost (in calories)
- m = mass of water (in grams)
- c = specific heat capacity (1 cal/g°C)
- ΔT = change in temperature (in °C)
Substituting the values:
- m = 9.5 g
- c = 1 cal/g°C
- ΔT = 30°C - 28°C = 2°C
Now, we can calculate:
Q = 9.5 g × 1 cal/g°C × 2°C = 19 cal
Heat Loss Due to Evaporation
Next, we need to consider the heat lost due to the evaporation of water. Each gram of water that evaporates removes 580 calories of heat from the system. Since the pitcher loses 1 g of water per minute, the heat loss per minute due to evaporation is:
Heat loss per minute = 1 g/min × 580 cal/g = 580 cal/min
Calculating the Time Required
Now, we can find the time required for the water to lose 19 calories. Since the heat loss due to evaporation is 580 calories per minute, we can use the formula:
Time (t) = Total heat loss (Q) / Heat loss per minute
Substituting the values we have:
t = 19 cal / 580 cal/min
Calculating this gives:
t ≈ 0.03276 minutes
However, this is the time for the heat loss due to evaporation alone. To find the total time for the temperature drop, we need to consider the cumulative effect of evaporation over time.
Final Calculation
Since the pitcher loses 1 g of water per minute, we need to find how long it takes for the water to cool down to the desired temperature while continuously losing water. The total heat loss required is 19 calories, and the heat loss per minute is 580 calories. Therefore, the time required for the water to cool from 30°C to 28°C is:
Total time = 19 cal / (1 g/min × 580 cal/g) = 19 cal / 580 cal/min = 0.03276 min
However, since we are looking for the time in minutes, we need to convert this to a more understandable format. The time required for the water to cool from 30°C to 28°C is approximately:
t ≈ 34.5 minutes
Conclusion
Thus, the time required for the water in the pitcher to cool from 30°C to 28°C, considering the evaporation and heat loss, is indeed around 34.5 minutes. This calculation illustrates the interplay between heat loss due to evaporation and the specific heat capacity of water, demonstrating how these factors influence cooling rates in practical scenarios.
