Condensation on a Glass and the Conservation of Heat
Problem Statement
Condensation on a Glass and the Conservation of Heat
Problem Statement
In a room, a piece of ice at an initial temperature of T1 was placed into a glass containing water of mass mw. After leaving it undisturbed for a while, the ice completely melted into water, and the temperature of all the water in the glass stabilized at T2.
From this state, the glass was left in the room for a further period of time, during which water vapor from the air condensed on the outside of the glass, forming water droplets with a mass of md. After a sufficient amount of time had elapsed, the water inside the glass and the water droplets on the outside combined to reach thermal equilibrium, and the overall temperature became T3. In this process, the following conditions are assumed:
・Heat transfer is considered only in terms of the latent heat released when water vapor condenses into liquid water on the outside of the glass; direct heat conduction from the surrounding air, the heat capacity of the glass itself, and the evaporation of water can be neglected. ・Let L be the amount of heat of condensation released when water vapor undergoes a phase transition to liquid water. ・Let cw be the specific heat of water.
Derive the temperature T3 at thermal equilibrium using an algebraic expression, then substitute the given constraint values to find the value of T3.
Constraints
- Mass of water in the glass: mw=0.158 kg
- Mass of ice added: mi=0.040 kg
- Initial temperature of the system: T2=278 K
- Mass of condensed water droplets: md=0.002 kg
- Latent heat of condensation of water: L=2.52×106 J/kg
- Specific heat of water: cw=4200J/(kg⋅K)
Input Format
Enter the value of the temperature T3 at thermal equilibrium as a natural number.
Solution
Explanation
Application of Physical Laws
Since we neglect direct heat conduction from the outside (the surrounding air) and the heat capacity of the container, the total energy of the system is conserved. The latent heat (heat of condensation) released when water vapor condenses on the outside of the glass is entirely used to raise the temperature of the water inside the glass and the water droplets adhering to it. Therefore, the law of conservation of heat holds true.
Deriving the Heat Changes for Each Substance
Let T3 be the overall temperature at thermal equilibrium.
- Heat Qrelease released by the condensation of water vapor This is the heat released when water vapor of mass md undergoes a phase transition and completely turns into liquid.
Qrelease=mdL
- Heat Qin absorbed by the water in the glass The glass contains water with a total mass (mw+mi) consisting of the original water mw and the ice mi that has melted into water. This water is heated from the initial temperature T2 to T3.
Qin=(mw+mi)cw(T3−T2)
- Heat Qout absorbed by the condensed water droplets The mass of the water droplets adhering to the outside is md. These droplets are also heated on the surface of the cup from the initial temperature T2 to the final temperature of the entire system, T3.
Qout=mdcw(T3−T2)
Derivation of T3 Using the Law of Conservation of Heat
Since the heat released is equal to the heat gained by the entire system, the equation for the conservation of heat is as follows:
Qrelease=Qin+Qout
mdL=(mw+mi)cw(T3−T2)+mdcw(T3−T2)
We factor out (T3−T2) from the right-hand side.
mdL=(mw+mi+md)cw(T3−T2)
We solve this equation for T3.
T3−T2=(mw+mi+md)cwmdL
T3=T2+(mw+mi+md)cwmdL
Substituting Values and Calculation
Substitute the given constraint values.
- Calculation of the numerator
mdL=0.002×2.52×106=5040 J
- Calculation of the denominator The total mass of the water being heated is as follows.
mw+mi+md=0.158+0.040+0.002=0.200 kg
The total heat capacity is given by:
(mw+mi+md)cw=0.200×4200=840 J/K
Substituting these values, we find the temperature change ΔT.
ΔT=8405040=6 K
Therefore, the final temperature T3 is as follows.
T3=278+6=284 K
The natural number to enter is 284.