Final temperature after pouring soda into glass with ice....

• NY152
In summary, the problem discusses the process of pouring soda from a 12 oz can at 13 degrees Celsius into a glass containing 0.16 kg of ice at -19.5 degrees Celsius. The specific heats of ice and water are given, as well as the latent heat of fusion of ice. The goal is to determine the final temperature of the mixture, how much of the ice has melted, and the mass of the melted ice. The solution involves setting up multiple equations to account for the phase changes that occur as the ice and water reach thermal equilibrium.
NY152

Homework Statement

Soda from a ms=12 oz can at temperature Ts=13 degrees C is poured in its entirety into a glass containing a mass mI=0.16kg amount of ice at temperature TI=-19.5 degrees C. Assume that ice and water have the following specfic heats c1=2090 J/(kg*C) and cs= 4186 J/(kg*C), and latent heat of fusion of ice is Lf=334 kj/kg.

a) In degrees celsius, what is the final temperature Tfinal of the mixture?
b) Write an expression for how much of the ice mmelted has melted?
c) In kilograms, how much of the ice mmelted has melted?

Q= m*c*deltaT

The Attempt at a Solution

So I think you start by converting the 12 oz can to kg to get the mass at that particular temperature. Then would you set two separate Q equations equal to one another to solve for Tf? So Q= ms*cs*(Tf-13) and Q= mI*cI*(Tf-(-19.5)) and set these equal to each other, but then how does the latent heat of fusion of ice play into the calculation?

Ok
NY152 said:

Homework Statement

Soda from a ms=12 oz can at temperature Ts=13 degrees C is poured in its entirety into a glass containing a mass mI=0.16kg amount of ice at temperature TI=-19.5 degrees C. Assume that ice and water have the following specfic heats c1=2090 J/(kg*C) and cs= 4186 J/(kg*C), and latent heat of fusion of ice is Lf=334 kj/kg.

a) In degrees celsius, what is the final temperature Tfinal of the mixture?
b) Write an expression for how much of the ice mmelted has melted?
c) In kilograms, how much of the ice mmelted has melted?

Q= m*c*deltaT

The Attempt at a Solution

So I think you start by converting the 12 oz can to kg to get the mass at that particular temperature. Then would you set two separate Q equations equal to one another to solve for Tf? So Q= ms*cs*(Tf-13) and Q= mI*cI*(Tf-(-19.5)) and set these equal to each other, but then how does the latent heat of fusion of ice play into the calculation?
O
NY152 said:

Homework Statement

Soda from a ms=12 oz can at temperature Ts=13 degrees C is poured in its entirety into a glass containing a mass mI=0.16kg amount of ice at temperature TI=-19.5 degrees C. Assume that ice and water have the following specfic heats c1=2090 J/(kg*C) and cs= 4186 J/(kg*C), and latent heat of fusion of ice is Lf=334 kj/kg.

a) In degrees celsius, what is the final temperature Tfinal of the mixture?
b) Write an expression for how much of the ice mmelted has melted?
c) In kilograms, how much of the ice mmelted has melted?

Q= m*c*deltaT

The Attempt at a Solution

So I think you start by converting the 12 oz can to kg to get the mass at that particular temperature. Then would you set two separate Q equations equal to one another to solve for Tf? So Q= ms*cs*(Tf-13) and Q= mI*cI*(Tf-(-19.5)) and set these equal to each other, but then how does the latent heat of fusion of ice play into the calculation?
Okay so I did: 4186*0.340194*(13-T) = 2090*0.16*(T+19.5)
and found Tf=6.82 and the system says this is incorrect but the hint says "You may have forgotten that ice experiences phase change at 0 degrees celsius. Take into account the latent heat.
As stated before, I'm not sure how to incorporate this into my solution.

I haven't gone through the numbers, but because of the wording of the question, I suspect this is what is happening in this problem.
The liquid gets added to the ice. The liquid warms up all of the ice to its melting point. Then the liquid melts all of the ice into a liquid. Then those two liquids reach some equilibrium temperature. Now it could be the case that the liquid does not have enough energy to do all of that. But that should come out in the numbers. So yes, you have to remember that the ice changes phase at 0C and requires energy from the warm(er) liquid to do that. And the amount of energy required to warm up 1 kg of ice 1 degree C is not the same amount of energy required to warm up 1 kg of water 1 degree C, which is not the same energy required to melt 1 kg of ice.

Edit: I defer to @Chestermiller. He is right (and I was wrong) that the problem statement seems to imply that not all of the ice is melted. I guess it would help if I read the whole problem carefully. :(

Last edited:
The problem statement (and your calculation) seems to imply that not all the ice will have melted when the system has reached thermal equilibrium. If that is the case, what is the final temperature of the system?

NY152 said:
Things will happen in stages, and one simple, if laborious, approach is to model those.
1. Ice warms and liquid cools until one of them reaches 0C.
2. If the water reaches 0C first, it will start to freeze, giving out latent heat.
2.1 If all the water freezes and the original ice is still less than 0C, the two will balance out at some temperature below freezing.
2.2 If the ice reaches 0C before all the water freezes you are done.
3. If the ice reaches 0C first then it will start to melt,
etc.

1. What factors affect the final temperature after pouring soda into a glass with ice?

The main factors that affect the final temperature are the initial temperature of the soda, the amount of ice, the temperature of the ice, and the type of soda being used.

2. Does the type of glass used affect the final temperature?

Yes, the type of glass can have an impact on the final temperature as different materials have different thermal conductivities, which can affect how quickly the soda cools down.

3. How does the temperature of the ice affect the final temperature?

The colder the ice, the more heat it will absorb from the soda, resulting in a lower final temperature. However, if the ice is too cold, it can cause the soda to freeze, which can affect the taste and carbonation.

4. Why does the soda sometimes become flat after pouring it into a glass with ice?

This can happen if the ice is not cold enough or if there is not enough ice to absorb all the heat from the soda. When this happens, the carbon dioxide in the soda can escape, causing it to become flat.

5. Can the final temperature be controlled?

Yes, the final temperature can be controlled to some extent by adjusting the initial temperature of the soda, the amount of ice, and the temperature of the ice. However, external factors such as the temperature of the surrounding environment can also play a role.

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