Water poured over ice until temp is at equilibrium

[castrodisastro]

Homework Statement

Suppose 337g of water at 32.9°C is poured over a 59.3g cube of ice with a temperature of -5.67°C. Does all the ice melt? If all the ice melts, what is the final temperature of the water? If all of the ice does not melt, how much ice remains when the ice-water mixture reaches equilibrium?

Specific Heat of water = 4.19*10³J/(kg*K)
Specific Heat of ice = 2.06*10³J/(kg*K)
Latent Heat of fusion of ice = 334*10³J/kg

Homework Equations

Q=mCΔT

The Attempt at a Solution

To determine if the process results in either ice, or ice-water mixture, I need to calculate how much heat is required to turn all the ice into water, then calculate how much heat was actually given off by pouring the water onto the ice.

So first

Heat required to raise the temperature of ice, and melt it completely
Q₁=m_iceC_iceΔT+m_iceL_{heat of fusion}

Q₁=(0.0593kg)(2.06*10³J/(kg*K))(5.67°C)+(0.0593kg)(334*10³)J/kg)
Q₁=20498.84J

20498.84 Joules required to raise the temp of the ice and melt it completely.

Now how much heat is given off when decreasing the temperature of the water to 0°C
Q₂=m_waterC_waterΔT

Q₂=(0.337kg)(4.19*10³J/(kg*K))(32.9°C)
Q₂=46455.79J

More heat is required to freeze the water than to melt the ice, so all the ice will melt.

Here is where I am stuck. So I need to find the final temperature of the water at equilibrium.
To melt the ice requires 20498.84 Joules so I set it equal to m_waterC_water(T_f-T_i) and solve for T_f.

This would tell me the temperature of the water right after the last bit of ice melted into water.

Rearranging this gives me

[Q₁+m_waterC_waterT_i]/[m_water]C_water = T_f

T_f=65.8°C (which is impossible since it can't get hotter than when it started)

I should be able to use T_f as the initial temperature of the next phase (rising temp of total water) heading towards equilibrium.

Am I not accounting for something? Should I not set
Q₁=m_waterC_waterΔT
to solve for T_final

Even after I do that, all I know is the temperature of the total water after the ice melted but have no way of knowing the final temp at equilibrium. If I want to use Q=mCΔT i have two unknows, T_final and Q, since I don't know how much heat is taken in as the mixture goes from colder freshly-melted ice-water + warmer liquid water to equilibrium.

What am I not understanding conceptually?

Thanks in advance.

P.S. Don't be rude please.

 

[SteamKing]

I follow your calculation of the heat required to melt the ice completely and the amount of heat available in the water which is poured over the ice.

Remember, once the ice has melted to liquid water, its mass mixes with that of the water being poured. You know how much heat is required for melting the ice. You now have to figure out the balance of heat afterwards.

You need to write an equation like this:

Q(melt water) gained = Q(pouring water) lost

 

[hjelmgart]

I am not sure how you rearranged it:

mwaterCwater(Tf-Ti)=Q1

(Tf-Ti)=Q1/mwaterCwater

With this you get the amount of kelvins, which the temperature of the water has decreased with (Tf-Ti) in order to melt the ice. You can then again use conservation of energy to do as suggested above, and remember to substract what you just found from the initial temperature of the water, and you should be able to find the equilibrium temperature.

 

[castrodisastro]

I follow your calculation of the heat required to melt the ice completely and the amount of heat available in the water which is poured over the ice.

Remember, once the ice has melted to liquid water, its mass mixes with that of the water being poured. You know how much heat is required for melting the ice. You now have to figure out the balance of heat afterwards.

You need to write an equation like this:

Q(melt water) gained = Q(pouring water) lost

So by Q_{melt water} you are referring to 20498.84 J right? Which is equal to the amount of heat the poured water lost when it melted the ice completely.

It should look like this
Q_{melt water}=m_{poured water}C_water(T_f-T_i)
correct?

The only value here that I don't know is the T_f so solving for it my equation becomes
Q_{melt water}=m_{poured water}C_waterT_f-m_{poured water}C_waterT_i

Q_{melt water}+m_{poured water}C_waterT_i=m_{poured water}C_waterT_f

(Q_{melt water}+m_{poured water}C_waterT_i)/m_{poured water}C_water=T_f

Now I plug in the values and solve for T_f

[(20498.84J)+(0.337kg)(4.19*10³J/(kg*K))(305.9K)]/[(0.337kg)(4.19*10³J/(kg*K))]=T_f

(20498.84J+431939.98J)/1412.03J/K)=T_f

320.417K=T_f

but 320.417K=47.4°C

The water started at 32.9°C, it can't possibly have gotten hotter. I know the math is correct so I am not using the correct information.

Am I correct to assume that for this last calculation, the T_f should be the temperature of the mixture when ice melts completely, and it is only at this time when I should consider the TOTAL mass of melted ice and poured water, and I should use this value to then determine that the T_f is as it goes from mixture of (poured water)+(freshly melted ice) to total water at equilibrium which is the next phase of this process?

 

[castrodisastro]

I am not sure how you rearranged it:

mwaterCwater(Tf-Ti)=Q1

(Tf-Ti)=Q1/mwaterCwater

With this you get the amount of kelvins, which the temperature of the water has decreased with (Tf-Ti) in order to melt the ice. You can then again use conservation of energy to do as suggested above, and remember to substract what you just found from the initial temperature of the water, and you should be able to find the equilibrium temperature.

Doing this my T_f came out to 320.41K

Which is higher than when it started. This is not possible.

T_f-T_i=(20498.84J)/(0.337kg)*(4.190*10³kJ/(kg*K))

T_f-T_i=14.517K

T_f=14.517K+T_i

T_f=14.517K+305.9K=320.41K

So instead I used the mass of the total water (melted ice cube+poured water)...

T_f-T_i=(20498.84J)/(0.3963kg)*(4.190*10³kJ/(kg*K))

T_f-T_i=12.35K

T_f=12.35K+T_i

T_f=12.35K+305.9K=318.24K

Still that temperature is higher which doesn't make sense.

Am I skipping something? Am I supposed to consider the heat Q₁ as negative? It is the only value that could possibly be negative to result in a decrease of the initial temperature. If I am cooling the water that began at 305.9K then is the heat considered negative because it is being lost?

Thanks

 

[hjelmgart]

Well what I meant, was to calculate ΔT and substract it from the initial. For some reason I was unsure, which was your Tf and Ti.. But if you calculate ΔT = Q1/m_wc_w you get 14.5K, then substract that from the starting temperature, as you know this value has to be negative, as the water transfers energy to the ice cube. So you get Tf = Ti-14.5K

But yeah a faster way is to just change the sign inside the paranthesis, since the energy has to be negative. You put Q1 = m_wc_wΔT but this is wrong, because Q1 has to be negative, when you are calculating it for the water, as it is the energy transferred from the water to the ice cube in order to melt it.

Else you can't have conservation of energy, and the 1st law is disobeyed.

 

[castrodisastro]

Well what I meant, was to calculate ΔT and substract it from the initial. For some reason I was unsure, which was your Tf and Ti.. But if you calculate ΔT = Q1/m_wc_w you get 14.5K, then substract that from the starting temperature, as you know this value has to be negative, as the water transfers energy to the ice cube. So you get Tf = Ti-14.5K

But yeah a faster way is to just change the sign inside the paranthesis, since the energy has to be negative. You put Q1 = m_wc_wΔT but this is wrong, because Q1 has to be negative, when you are calculating it for the water, as it is the energy transferred from the water to the ice cube in order to melt it.

Else you can't have conservation of energy, and the 1st law is disobeyed.

Ok makes sense. So my final temperature is 305.9K-14.5K=291.38K when the ice has fully melted. The mixture however has not necessarily reached equilibrium.

Now the last part should be the part of the process that goes from 291.38K to equilibrium correct? We shall call it, let's say, Q₃

Q₃=m_{total water}C_water(T_f-T_i)

But here I have two unknowns, Q₃ and T_f

Up to this point 20498.84 Joules have been lost by the poured water as it gets colder, but since it is an open system, more heat is lost to the surroundings as the mixture cools from 291.38K to equilibrium. Since this is the case, I don't have a specific value for heat that I can equate to Q₃ so that I can solve the above problem. How can I know what amount to use for Q to be able to solve T_equilibrium?

 

[hjelmgart]

I think SteamKing already answered this last question of yours, and if you think more about it, it's quite obvious.

You are now left with two substances of water, one at 291K and one at 273K, they have different mass, but same heat capacity. So you want to find an equilibrium temperature between each quantity, you call it Q₃, which is fine.

Now remember, that there has to be conservation of energy. The energy transferred from the water at 291K has to be equal to energy absorbed by the water at 273K, for both fluids to be in equilibrium.

Try to set up two equations for Q₃ and put them equal to each other, and see if it can then be solved, one for each quantity of water.

 

[castrodisastro]

I think SteamKing already answered this last question of yours, and if you think more about it, it's quite obvious.

You are now left with two substances of water, one at 291K and one at 273K, they have different mass, but same heat capacity. So you want to find an equilibrium temperature between each quantity, you call it Q₃, which is fine.

Now remember, that there has to be conservation of energy. The energy transferred from the water at 291K has to be equal to energy absorbed by the water at 273K, for both fluids to be in equilibrium.

Try to set up two equations for Q₃ and put them equal to each other, and see if it can then be solved, one for each quantity of water.

Ok I see. So I would do

Q₃=m_{poured water}C_water(T_f-T_i)

should be equal to (due to conservation of energy)

Q₃=m_{melted ice}C_water(T_f-T_i)

I should set them equal to each other and solve for T_f.

m_{poured water}C_water(T_f-T_i)=m_{melted ice}C_water(T_f-T_i)

I can cancel out the heat capacity since it is on both sides...

multiplying out gives me this

m_{poured water}T_f-m_{poured water}T_i=m_{melted ice}T_f-m_{melted ice}T_i

Solving for T_f

m_{poured water}T_f-m_{melted ice}T_f=m_{poured water}T_i-m_{melted ice}T_i

Factor the T_f out

T_f(m_{poured water}-m_{melted ice})=m_{poured water}T_i-m_{melted ice}T_i

Finally

T_f=[(m_{poured water}T_i-m_{melted ice}T_i)/(m_{poured water}-m_{melted ice})]

That should give me the correct answer. I see how conservation of energy works here, but is this a correct translation of the concept into calculation of it?

 

[SteamKing]

So by Q_{melt water} you are referring to 20498.84 J right? Which is equal to the amount of heat the poured water lost when it melted the ice completely.

I should have been more clear in my original post.

The Q = 20498.84 J has already been used to melt the ice and turn it to liquid water at 0 C.

There is an unknown final temperature Tf where the melt water and the poured water reach thermal equilibrium. It is for this circumstance that you write the equation:

Q(gained by melt water) = Q(lost by poured water)

 

[castrodisastro]

I should have been more clear in my original post.

The Q = 20498.84 J has already been used to melt the ice and turn it to liquid water at 0 C.

There is an unknown final temperature Tf where the melt water and the poured water reach thermal equilibrium. It is for this circumstance that you write the equation:

Q(gained by melt water) = Q(lost by poured water)

Ok thanks yes I understood now what you were trying to tell me, so the calculations in my previous post are still correct right? I only needed the Q = 20498.84 J to determine the change in temperature of the poured water at 32.9 °C to allow me to calculate Q₃ as jhelmgart explained?

 

[hjelmgart]

Well I might not have expressed myself clearly before. You did the math correctly, but remember that when a substance transfers heat (or loses heat) it has to be a negative value.

 

[castrodisastro]

Well I might not have expressed myself clearly before. You did the math correctly, but remember that when a substance transfers heat (or loses heat) it has to be a negative value.

So in my last post I have arrived at the final temperature correctly? Just double checking before I fully submit my assignment to check my score.

Thanks again for the help. Much appreciated

 

[hjelmgart]

I don't think so. You have to do put a minus sign in front of Q_poured, or else they won't be equal. I think it would change the sign in the upper and lower bracket of your paranthesis?

You would probably get a temperature that doesn't make sense the way you did it?

 

[castrodisastro]

ok so what I should have done was...

-m_{poured water}(T_f-T_i)=m_{melted ice}(T_f-T_i)

So the Q for the poured water should be negative since that process involves losing heat(getting colder) and the melted ice is positive because it is gaining heat (getting warmer).

 

[castrodisastro]

the reason that is true is because

To clear up a conceptual misunderstanding what about the fact that

Q_total=Q_ice+Q_water

Q_ice will increase in temperature and it will take in heat. This value will be positive.

Q_water will decrease in temperature and it will lose heat. This value will be negative.

If they reach equilibrium, then the magnitudes will be equal to each other.

Q_total=Q_ice+Q_water=0

This means that 0=(a positive x value)+(a negative x value)

If 0=(x)+(-x)

x=x

How could I make sense of a negative x value being equal to a positive x value?

 

[hjelmgart]

But Q_water itself, when you calculate it, gives a negative value, because the temperature T_f is smaller than T_i. This isn't the case for Q_ice.

So they wouldn't be equal to each other unless you put the negative sign in front of Q_water.

In other words Q_tot = Q_ice + Q_water = 0, you then isolate it to be Q_ice = -Q_water, hence the negative sign.

Perhaps I was a bit unclear or said something wrong. Anyway this reasoning is probably better :) good luck

 

[castrodisastro]

Got it! Thanks!