Liquid water boils at 100 degrees C. In other words, this is the temperature at which the thermal energy between two molecules of water overcomes the interaction energy. Given this, what is the interaction energy (Uint) of liquid water?
3/2*(1.38*10e-23)*373.15= 7.7e-20
7.72*10-21J
To find the interaction energy (Uint) of liquid water, we can use the formula:
Uint = 3/2 * k * T
Where:
- k is the Boltzmann constant (approximately 1.38 * 10^-23 J/K)
- T is the temperature in Kelvin
Since water boils at 100 degrees Celsius, we need to convert it to Kelvin by adding 273.15:
T = 100 + 273.15 = 373.15 K
Now, we can substitute the values into the formula:
Uint = 3/2 * (1.38 * 10^-23 J/K) * 373.15 K ≈ 7.7 * 10^-20 J
So, the interaction energy (Uint) of liquid water is approximately 7.7 * 10^-20 J.