How many kilo calories (kcal) are required to change 1 kg of 0°C water to 1 kg of 100°C water?
To calculate the amount of energy required to change the temperature of water from 0°C to 100°C, we need to consider two separate steps: raising the temperature of the water from 0°C to 100°C and then changing the state of the water from liquid to gas at its boiling point.
First, let's determine the energy required to raise the temperature of water from 0°C to 100°C. The specific heat capacity of water is approximately 4.18 Joules per gram per degree Celsius (J/g°C). Since there are 1000 grams in 1 kilogram, we can use this value to calculate the energy required to heat up the water:
Energy = Mass × Specific Heat Capacity × Temperature Change
Mass of water = 1000 grams
Specific heat capacity of water = 4.18 J/g°C
Temperature change = 100°C - 0°C = 100°C
Energy = 1000 g × 4.18 J/g°C × 100°C
Energy = 418,000 J
Now, let's consider the energy required to change the state of water from liquid to gas at its boiling point (100°C). This is known as the heat of vaporization. The heat of vaporization for water is approximately 2260 J/g.
Energy = Mass × Heat of Vaporization
Mass of water = 1000 grams
Heat of vaporization of water = 2260 J/g
Energy = 1000 g × 2260 J/g
Energy = 2,260,000 J
Finally, we add up the energy required to raise the temperature and the energy required for vaporization:
Total Energy = Energy to raise the temperature + Energy for vaporization
Total Energy = 418,000 J + 2,260,000 J
Total Energy = 2,678,000 J
Since 1 kilocalorie (kcal) is equal to 4,184 J, we can convert the total energy from joules to kilocalories:
Total Energy (in kcal) = 2,678,000 J / 4,184 J/kcal
Total Energy (in kcal) ≈ 639.17 kcal
Therefore, approximately 639.17 kilocalories are required to change 1 kg of 0°C water to 1 kg of 100°C water.