On a certain dry sunny day, a swimming pool's temperature would rise by 1.60°C if not for evaporation. What fraction of the water must evaporate to carry away precisely enough heat to keep the temperature constant?
(Lv =2430kJ/kg or 580kcal/kg)
To calculate the fraction of water that must evaporate to maintain a constant temperature, we need to understand the heat energy involved in the process of evaporation.
The heat energy required for the evaporation of water can be calculated using the formula: Q = m * Lv, where Q is the heat energy, m is the mass of water evaporated, and Lv is the latent heat of vaporization.
Given that Lv = 2430 kJ/kg or 580 kcal/kg, we can use this information to find the heat energy required to maintain a constant temperature.
First, let's assume the swimming pool's temperature is initially at the desired constant value. If no evaporation occurs and the temperature rises by 1.60°C, we know that the heat energy gained by the water is equal to the heat energy required to maintain the temperature constant.
The heat energy gained (Q_gained) can be calculated using the formula: Q_gained = m * c * ΔT, where m is the mass of water, c is the specific heat capacity of water (which is approximately 4.18 kJ/(kg·°C)), and ΔT is the change in temperature.
In this case, the change in temperature is 1.60°C.
Now, to find the mass of water that needs to evaporate to carry away precisely enough heat to keep the temperature constant, we equate the heat energy gained to the heat energy required for evaporation.
m * c * ΔT = m * Lv
We can cancel out the mass (m) on both sides of the equation, resulting in:
c * ΔT = Lv
Plugging in the values, we have:
(4.18 kJ/(kg·°C)) * 1.60°C = 2430 kJ/kg
Now, we can calculate the fraction of water that needs to evaporate. Let's assume the initial mass of the water is 1 kg.
Fraction of water evaporated = (mass evaporated) / (initial mass)
To find the mass evaporated, we solve for it in the equation:
mass evaporated = Fraction of water evaporated * initial mass
mass evaporated = (2430 kJ/kg) / (4.18 kJ/(kg·°C)) * 1.60°C
mass evaporated = 910.05 g or 0.91005 kg
Fraction of water evaporated = (0.91005 kg) / (1 kg)
The fraction of water that must evaporate to carry away enough heat to keep the temperature constant is approximately 0.91005.
To determine the fraction of water that must evaporate to maintain a constant temperature, we need to calculate the amount of heat that is being carried away by evaporation.
The heat required to change the temperature of water can be calculated using the specific heat capacity:
Q = m * c * ΔT
Where:
Q = heat energy (in joules)
m = mass of water (in kg)
c = specific heat capacity of water (in joules per kilogram per degree Celsius)
ΔT = change in temperature (in degrees Celsius)
Using the given values:
ΔT = 1.60°C
c = 4186 J/kg°C (specific heat capacity of water)
Now we can solve for Q.
Q = m * c * ΔT
To find out the mass of water (m) that needs to evaporate, we need to convert Q into calories and divide it by the heat of vaporization (Lv).
1 kcal = 4186 J
1 kg of water = 1 liter of water = 1000 grams
To find the mass in grams, we divide the calculated heat energy Q by Lv, and then convert it to kilograms.
Here are the steps:
1. Convert Lv to joules:
Lv = 2430 kJ/kg = 2430 * 1000 J/kg = 2,430,000 J/kg
2. Convert Q (heat energy) to calories:
Q = m * c * ΔT
Q = 1 kcal (since we want Q in calories)
1 kcal = 4186 J
Q = Q * 4186
3. Find the mass of water that needs to evaporate:
m = Q / Lv
m = (Q * 4186) / 2,430,000
4. Convert mass from grams to kg:
m = m / 1000
Finally, we can express the mass evaporation as a fraction of the total mass of water in the swimming pool. Let's assume the total mass of water in the pool is M.
Fraction = mass of water evaporated / total mass of water
Fraction = m / M
Please provide the total mass of water (M) in the swimming pool to finalize the calculation.
Q=Mc ΔT =mL(v)
m/M = c ΔT/L9v) =
=4183•1.6/2430000 = 0.00275
or 0.275%