To tackle this problem, we need to analyze the heat loss due to evaporation and how it affects the temperature of the water in the earthen pot. The key concepts here involve the heat of vaporization, the mass loss over time, and the relationship between heat transfer and temperature change. Let's break it down step by step.
Understanding the Problem
We have an earthen pot containing 10 kg of water, and water is leaking out at a rate of 0.2 grams per second. As this water evaporates, it takes away heat from the pot, leading to a decrease in temperature. Our goal is to find out how long it will take for the temperature of the water to drop by 5 °C.
Key Concepts
- Heat of Vaporization: This is the amount of energy required to convert a unit mass of a liquid into vapor without a change in temperature. For water, this value is approximately 2260 kJ/kg.
- Specific Heat Capacity: This is the amount of heat required to raise the temperature of a unit mass of a substance by 1 °C. For water, it is about 4.18 kJ/kg°C.
Calculating Heat Loss
First, we need to calculate the total heat loss required to decrease the temperature of the water by 5 °C. We can use the formula:
Q = mcΔT
Where:
- Q = heat energy (in kJ)
- m = mass of water (in kg)
- c = specific heat capacity of water (4.18 kJ/kg°C)
- ΔT = change in temperature (in °C)
Substituting the values:
Q = 10 kg × 4.18 kJ/kg°C × 5 °C
Calculating this gives:
Q = 209 kJ
Heat Loss Due to Evaporation
Next, we need to determine how much heat is lost due to the evaporation of water. Since 0.2 grams of water evaporates per second, we can convert this to kilograms:
0.2 g/s = 0.0002 kg/s
The heat loss per second due to evaporation can be calculated using:
Heat loss per second = mass loss per second × heat of vaporization
Substituting the values:
Heat loss per second = 0.0002 kg/s × 2260 kJ/kg
This results in:
Heat loss per second = 0.452 kJ/s
Time Calculation
Now, we can find out how long it will take to lose the required 209 kJ of heat. We use the formula:
Time = Total heat loss required / Heat loss per second
Substituting the values:
Time = 209 kJ / 0.452 kJ/s
Calculating this gives:
Time ≈ 462.6 seconds
To convert this into minutes:
Time ≈ 462.6 seconds / 60 ≈ 7.7 minutes
Final Thoughts
Thus, it will take approximately 7.7 minutes for the temperature of the water in the pot to decrease by 5 °C due to the evaporation of water. This problem illustrates the interplay between mass loss, heat transfer, and temperature change, showcasing the principles of thermodynamics in a practical scenario.
