# Thermal Equillibrium (I am missing something here)

1. Nov 17, 2007

This problem is giving me great fun:

Problem A person makes iced tea by mixing 500 g of hot tea (essentially water) with an equal mass of ice at its melting point. Assume the mixture has negligble energy exchanges with its environment. If the teas initial tempurature is T_i=90 deg celsius...when thermal equillibrium is reached what are (a) the mixtures final tempurature T_f and the remaining mass m_f of the ice?

Relevant Eqs Since the mixture has negligble energy exchanges with its environment, $$\sum Q=0$$

$$Q=mc\Delta T$$ and $$Q=mL_f$$

My crappy reasoning skills

So I initially have: $$m_{w}=500g @ 90^{\circ}\stackrel{Q_1=(mc_w\Delta T)}{\rightarrow} T_f$$

$$m_2= 500g@0^{\circ}\stackrel{Q_2=m_2L_f}{\rightarrow}T_f$$

and I think I need a 3rd Q where :$$Q_3=(mc\Delta T)$$

So I have 500g Water at 90 C--->T_f
500g Ice at 0 C----> some mass of water at 0 C
and then some mass of Water at 0 C--->T_f

Here is where I am getting all confused. Can anyone help point me in the right direction?

Casey

2. Nov 17, 2007

### dynamicsolo

It's a good idea to start off by checking whether there is enough heat available from the tea to melt all the ice. The calorimetry equation will change a bit if there is ice left over at equilibrium. How much heat is needed to completely melt 500 g of ice? What would the temperature of tea by lowered to by surrendering that much heat?

If the answer is above 0º C., you can then look for the equilibrium between 500 g of meltwater at 0º C. and 500 g of tea starting at the reduced temperature.

If the answer is less than or equal to 0º C., you will instead need to find how much ice can be melted by having the tea give up the heat associated with a 90º C. temperature reduction. The equilibrium temperature will be 0º C.

3. Nov 17, 2007

Well, from the question, it is implied that there is ice left over...this maybe why my professor did not actually assign this problem. I just came across it in the text.

How does the method differ? I would not mind learning it. Does it require an itegration?

Casey

4. Nov 17, 2007

### katchum

It probably doesn't acquire integration. With good mixing the temperature will stay at 0°C.

Last edited: Nov 17, 2007
5. Nov 17, 2007

### dynamicsolo

An integration is "overkill", since the relations are all linear.

The heat released by the tea in going from 90º C to 0º C would be

Q_cool = (m_tea)·(c_water)·(T_f - T_i) = (500 g)(1 cal/g-C)(0º - 90º C) = -45,000 cal ,

while the heat required to melt 500 g of ice at 0º C is

Q_melt = (m_ice)·(L_f) = (500 g)(79.8 cal/g) = 39,900 cal.

So there is more than sufficient hot tea to melt all the ice. (Questions of this sort will ask for the mass of ice remaining even when the answer is zero. Had there been 450 g of hot tea or had it been 500 g initially at 75º C, there would be a different story...)

We now have 500 g of meltwater at 0ºC mixed with 500 g of cooled tea. The tea will have transferred 39,900 calories of heat to the ice, so its temperature will have already reduced to

(Q_cool)' = (500 g)(1 cal/g-C)(T_f - 90) = -39,900 cal

which gives (T_f - 90) = -79.8 [not surprising, given how we got the 39,900 in this problem] or T_f = 10.2º C.

The heat transfer between the meltwater and the tea is now

(m_water)(c_water)(T_f' - 0º C) = -(m_tea)(c_water)(T_f' - T_f)

or

(500 g)(1 cal/g-C)(T_f' - 0º) = -(500 g)(1 cal/g-C)(T_f' - 10.2º)

which gives 1000 T_f' = 5105 or T_f' = 5.11º C. (Since the masses of meltwater and tea are equal in this problem, and their specific heat capacities are taken to be the same, you could just average 0º and 10.2º .)

Last edited: Nov 17, 2007
6. Nov 17, 2007

Thanks ds. I like the simple "check" that you did. I don't know why I did not think of that :/ I am going to rework this problem using your method and see if I can adapt it to my reasoning.

Thanks,
Casey

7. Nov 17, 2007

### dynamicsolo

Most textbooks I've seen will illustrate such a problem by writing the heat transfer equation as

Q_melt + Q_warm = -Q_cool , giving

(m_ice)(L_f) + (m_meltwater)(c_water)(T_f - 0º C) = -(m_tea)(c_water)(T_f - 90º C) .

This is fine in the case where all of the ice melts, so that you're just using the heat from the tea to melt the ice and warm up the meltwater. But, inevitably, there is a homework or exam problem where all the ice doesn't melt, in which case you run into a problem getting the equilibrium temperature from this equation (you get a result below 0º C). My approach to heading this off is to check on the available heat to transfer first...

Last edited: Nov 17, 2007
8. Nov 19, 2007