What is the temperature of a system in thermal equilibrium with another system made up of water and steam at one atmosphere of pressure? a. 0°F b. 273 K c. 0 K d. 100°C
The correct answer and explanation is:
Correct Answer: d. 100°C
When a system is in thermal equilibrium with a mixture of liquid water and steam at one atmosphere (1 atm) of pressure, its temperature is 100 degrees Celsius (°C).
At 1 atm of pressure, water boils at 100°C. This temperature remains constant during the phase transition between liquid and vapor. Even though energy is continuously added to convert the water into steam, the temperature does not increase until the entire liquid has turned into vapor. This is a key property of phase transitions at constant pressure.
Thermal equilibrium means two systems in contact do not exchange net heat energy because they are at the same temperature. So, if one system is in contact with water-steam mixture at 1 atm and reaches thermal equilibrium, it must be at exactly 100°C, the boiling point of water under those conditions.
This principle is a foundational concept in thermodynamics. It is used in calibration of thermometers and other temperature-sensing instruments. The triple point and boiling point of water serve as reliable fixed reference temperatures.
Now let’s evaluate the incorrect choices:
- a. 0°F is approximately -17.78°C, which is much colder than boiling water and not in equilibrium with steam at atmospheric pressure.
- b. 273 K is equal to 0°C, the freezing point of water, not the boiling point.
- c. 0 K is absolute zero, the lowest possible temperature, where all molecular motion ceases. It is physically impossible to reach and certainly not in equilibrium with a boiling mixture.
Therefore, the only correct temperature for a system in thermal equilibrium with a boiling water-steam mixture at 1 atm is 100°C.