Which requires more energy: to raise 1 kg of liquid water from 0 to 100 degrees (Celsius) or to turn that water to steam after it hits 100 degrees? Assume the specific heat capacity of water is approximately independent of temperature.
Heating of the liquid from 0 to 100 C requires 100 cal/g. Evaporation requires 540 cal/g, more than five times as much heat.
To determine which process requires more energy, we need to consider the heat capacity and the latent heat of vaporization.
1. Raising the temperature of water from 0 to 100 degrees Celsius:
The heat capacity of water is approximately 4.18 Joules/gram°C, which means it takes 4.18 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius. Since we have 1 kg of water, we multiply the heat capacity by 1000 to get the total heat capacity.
Total heat capacity = (4.18 J/g°C) x (1000 g) = 4180 Joules/°C
To raise the temperature from 0 to 100 degrees Celsius, we need to multiply the total heat capacity by the temperature range:
Energy required = Total heat capacity x Temperature range
= 4180 J/°C x (100 °C - 0 °C)
= 4180 J/°C x 100 °C
= 418,000 Joules
Therefore, it takes 418,000 Joules of energy to raise 1 kg of liquid water from 0 to 100 degrees Celsius.
2. Turning water to steam after reaching 100 degrees Celsius:
The latent heat of vaporization for water is approximately 2,260 Joules/g. Since we have 1 kg of water, we multiply the latent heat by 1000 to get the total latent heat of vaporization.
Total latent heat = (2,260 J/g) x (1000 g) = 2,260,000 Joules
Therefore, it takes 2,260,000 Joules of energy to turn 1 kg of liquid water to steam after it reaches 100 degrees Celsius.
Comparing the two processes, we can see that turning water to steam requires significantly more energy than raising its temperature from 0 to 100 degrees Celsius.
To determine which requires more energy, let's break down the process and calculate the energy required for each step separately.
First, let's calculate the energy required to raise the temperature of 1 kg of liquid water from 0 to 100 degrees Celsius. We can use the formula:
Q = m * c * ΔT
Where:
- Q is the energy required (in joules)
- m is the mass of the water (1 kg)
- c is the specific heat capacity of water (approximately 4,186 J/kg°C)
- ΔT is the change in temperature (100°C - 0°C = 100°C)
Plugging in the values, we get:
Q1 = 1 kg * 4,186 J/kg°C * 100°C
Q1 = 418,600 joules
So, it requires 418,600 joules of energy to raise 1 kg of liquid water from 0 to 100 degrees Celsius.
Next, let's calculate the energy required to turn the 1 kg of water at 100 degrees Celsius into steam. The energy required for this phase change is called the heat of vaporization. For water, the heat of vaporization is approximately 2,260,000 joules/kg.
Q2 = m * Lv
Where:
- Q2 is the energy required for vaporizing the water (in joules)
- m is the mass of the water (1 kg)
- Lv is the heat of vaporization of water (approximately 2,260,000 J/kg)
Plugging in the values, we get:
Q2 = 1 kg * 2,260,000 J/kg
Q2 = 2,260,000 joules
So, it requires 2,260,000 joules of energy to turn 1 kg of water at 100 degrees Celsius into steam.
Comparing the two calculations, we can conclude that it requires more energy to turn the water into steam after it has reached 100 degrees Celsius. Specifically, it requires 2,260,000 joules to turn the water into steam, compared to 418,600 joules needed to raise its temperature from 0 to 100 degrees Celsius.