Calculate the entropy change of the water while it cools

In summary, the conversation discusses the calculation of entropy change for water cooling from 85.0 C to 20.0 C and the air in the kitchen remaining isothermal. The solution for part A is provided, but the individual is unsure about their approach for part B. They attempt to use the equation S=Q/T, but use the latent heat of fusion for water instead of the correct value. The correct approach is to calculate the heat flow for the air, -Qwater, and divide by the temperature to get the change in entropy of the air.
  • #1
anubis01
149
1

Homework Statement



You make tea with 0.250 Kg of 85.0 C water and let it cool to room temperature (20.0 C) before drinking it.

a)Calculate the entropy change of the water while it cools.
b)The cooling process is essentially isothermal for the air in your kitchen. Calculate the change in entropy of the air while the tea cools, assuming that all the heat lost by the water goes into the air.
c)What is the total entropy change of the system tea + air?

Homework Equations


S=Q/T
Q=mLf
S=mcwln(T2/T1)



The Attempt at a Solution


I solved part A but I'm having trouble with part b.

a) S=mcwln(T2/T1)=0.25*4190(ln(293.15/358.15))
=-210J/K

b) Since its isothermal I use the equation S=Q/T. T=20+273.15
S=Q/T=mLf/T=(0.25*3.34x10^5)/(293.15)=285 J/K

I'm not sure if what I did for part b is correct, so if someone could tell me if I'm on the right track I would appreciate it. Thanks for the help.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2


anubis01 said:

Homework Equations


S=Q/T
Q=mLf
S=mcwln(T2/T1)

The Attempt at a Solution


I solved part A but I'm having trouble with part b.

a) S=mcwln(T2/T1)=0.25*4190(ln(293.15/358.15))
=-210J/K

b) Since its isothermal I use the equation S=Q/T. T=20+273.15
S=Q/T=mLf/T=(0.25*3.34x10^5)/(293.15)=285 J/K

I'm not sure if what I did for part b is correct, so if someone could tell me if I'm on the right track I would appreciate it. Thanks for the help.
Where do you get 3.34 x 10^5? You appear to be using the latent heat of fusion for water, which has no application here at all.

The heat flow, Qair is simply -Qwater. Work that out and divide by 293K to get the change in entropy of the air.

AM
 
  • #3


Your solution for part a is correct. For part b, you are on the right track but your calculation is not entirely correct. The heat lost by the water does not go directly into the air, as some of it may be lost to the surroundings. Therefore, we need to consider the total heat lost by the water, which is equal to the heat gained by the air. This can be calculated using the following equation:

Q = mCΔT

Where Q is the heat gained by the air, m is the mass of the air, C is the specific heat capacity of air, and ΔT is the change in temperature of the air (which is the same as the change in temperature of the water).

We can rearrange this equation to solve for the change in entropy of the air:

S = Q/T = (mCΔT)/T

Substituting in the values, we get:

S = (0.25*1005*(85-20))/(293.15) = 71.6 J/K

Therefore, the change in entropy of the air is 71.6 J/K.

For part c, we simply need to add the change in entropy of the water (from part a) and the change in entropy of the air (from part b):

S_total = S_water + S_air = -210 J/K + 71.6 J/K = -138.4 J/K

Therefore, the total entropy change of the system (tea + air) is -138.4 J/K.
 

FAQ: Calculate the entropy change of the water while it cools

1. What is entropy and why is it important to calculate its change in water cooling?

Entropy is a measure of the disorder or randomness in a system. In the context of water cooling, it represents the change in the distribution of thermal energy as the water loses heat. It is important to calculate entropy change in order to understand and predict the behavior of the water during the cooling process.

2. How is the entropy change of water calculated?

The entropy change of water while cooling can be calculated using the equation ΔS = Q/T, where ΔS is the change in entropy, Q is the heat transferred, and T is the temperature at which the heat transfer occurs. This equation is based on the Second Law of Thermodynamics, which states that the total entropy of a closed system always increases.

3. What factors affect the entropy change of water during cooling?

The entropy change of water during cooling is affected by several factors, including the temperature difference between the water and its surroundings, the heat capacity of water, and the amount of heat transferred. Additionally, the presence of impurities or dissolved substances in the water can also affect the entropy change.

4. Can the entropy change of water during cooling be negative?

Yes, the entropy change of water during cooling can be negative. This occurs when the temperature of the water increases, causing the randomness or disorder of the molecules to decrease. In this case, the heat transfer is from the colder surroundings to the warmer water, which goes against the natural tendency of energy to flow from hot to cold.

5. How is the entropy change of water related to its phase change during cooling?

The entropy change of water is closely related to its phase change during cooling. When water changes from a liquid to a solid (freezing), the entropy decreases because the molecules become more ordered in a solid state. On the other hand, when water changes from a liquid to a gas (evaporation), the entropy increases due to the increased randomness of the gas molecules. These phase changes also affect the temperature at which the entropy change occurs.

Back
Top