A 5000 kg African elephant has a resting metabolic rate of 2500 W. On a hot day, the elephant's environment is likely to be nearly the same temperature as the animal itself, so cooling by radiation is not effective. The only plausible way to keep cool is by evaporation, and elephants spray water on their body to accomplish this.
If this is the only possible means of cooling, how many kilograms of water per hour must be evaporated from an elephant's skin to keep it at a constant temperature?
Express your answer using two significant figures.
If you could provide an explanation that would be great.
To calculate the amount of water that needs to be evaporated from an elephant's skin to keep it at a constant temperature, we need to use the concept of specific latent heat of vaporization. The specific latent heat of vaporization is the amount of heat energy required to change a substance from liquid to gas at a constant temperature and pressure.
1. First, we need to calculate the heat energy that needs to be removed from the elephant's body to maintain its constant temperature. This heat energy is equal to the resting metabolic rate of the elephant, which is given as 2500 W.
2. We can then convert this heat energy into joules per hour. Since 1 Watt is equal to 1 Joule per second, we can multiply the resting metabolic rate by the number of seconds in an hour (3600 seconds) to get the heat energy in joules per hour:
2500 W * 3600 s = 9,000,000 J/hour
3. Next, we need to determine the specific latent heat of vaporization of water, which is the amount of heat energy required to evaporate 1 gram of water. The specific latent heat of vaporization of water is approximately 2.26 x 10^6 J/kg.
4. We can now use the equation to find the mass of water that needs to be evaporated:
Mass of water (kg) = Heat energy (J) / Specific latent heat of vaporization (J/kg)
Mass of water (kg) = 9,000,000 J / 2.26 x 10^6 J/kg
Calculating this gives us:
Mass of water (kg) ≈ 3.98 kg
Therefore, approximately 4 kg of water per hour must be evaporated from an elephant's skin to keep it at a constant temperature.
To determine the amount of water that needs to be evaporated from an elephant's skin to keep it at a constant temperature, we need to consider the energy required for the evaporation process.
The energy required to evaporate a certain amount of water can be calculated using the formula:
Energy (E) = Mass (m) x Latent heat of vaporization (L)
Where:
- Energy (E) is measured in Joules (J)
- Mass (m) is the amount of water evaporated, measured in kilograms (kg)
- Latent heat of vaporization (L) is the amount of energy required to evaporate 1 kilogram of water at a specific temperature, measured in Joules per kilogram (J/kg)
The latent heat of vaporization of water at room temperature is approximately 2,260,000 J/kg.
Now, let's calculate the energy required to maintain the elephant's temperature:
Power (P) = 2500 W (given in the problem)
Time (t) = 1 hour
The energy required to maintain the temperature can be obtained by multiplying the power by the time:
Energy required = Power x Time
Energy required = 2500 W x 1 hour = 2500 Joules
Since the only means of cooling for an elephant is through evaporation, this energy needs to be balanced by the energy required to evaporate water. Therefore, we have:
Energy required = Energy for evaporation
2500 J = m (kg) x 2,260,000 J/kg
To find the mass of water that needs to be evaporated (m), we rearrange the equation:
m = (2500 J) / (2,260,000 J/kg)
m ≈ 0.00111 kg
Therefore, approximately 0.00111 kg (or 1.11 grams) of water per hour must be evaporated from an elephant's skin to maintain its constant temperature.
Note: The given problem assumes that all the energy is consumed in evaporation, neglecting any losses due to other factors like heat radiation or conduction.
heat loss=2500watts
masswater/time*Heatvaporization=2500
solve for masswater per second.
Now, convert that to kg/hr