Intro Thermodynamics-put ice in water

1. Jan 31, 2007

mbrmbrg

Intro Thermodynamics--put ice in water

1. The problem statement, all variables and given/known data

(a) Two 45 g ice cubes are dropped into 290 g of water in a thermally insulated container. If the water is initially at 25°C, and the ice comes directly from a freezer at -15°C, what is the final temperature at thermal equilibtrium? (Neglect the heat capacity of the glass.)

(b) What is the final temperature if only one ice cube is used?

2. Relevant equations

Q_out=Q_in

For temperature change without phase change:
$$Q=cm\Delta T$$

For phase change:
$$Q=Lm$$

oC + 273.15 = K

3. The attempt at a solution

I solved part a; as the energy needed to melt the ice is greater than the heat available from the water the system stabilizes at the melting/freezing point of water.

For part b, to check if the water has enough energy to melt the ice, I computed (where the supscript i stands for ice) $$c_im_i\Delta T + L_im_i = (2220 \frac{J}{kg K})(0.045 kg)(288.15 K - 273.15 K) + (333000 \frac{J}{kg})(0.045 kg) = 16483.5 J$$

I had found in part a that the energy that the water can give off before freezing is $$c_wm_w \Delta T = (4190 \frac{J}{kg K})(0.290 kg)(298.15) = 30377.5 J$$.

Since Q_i < Q_w, the ice will melt completely and the water temperature will rise. Thus, my equation for part b ought to read as follows:

$$c_im_i(T_f-T_{ice initial}) + L_im_i = -c_wm_w(T_f-T_{water initial})$$
Yes, I should have done algebra here, but I lazily chose (and continue to choose) not to; I just plug and play as long as I can. Also, since we're always dealing with Delta T rather than T proper, converting to Kelvins will end up cancelling, so I left everything in Celcius (which is what the problem asks for, anyways). So:

$$(2220)(0.045)(T_f - (-15)) + (333000)(0.045) = -(4190)(0.290)(T_f - 25)$$

This arithmetics out to $$99.9T_f + 1498.5 = -1215.5T_f + 30377.5$$

Get all the T_f's onto one side, divide by their coefficient, and get T_f = 21.96 degrees Celcius.

Then WebAssign says that that is the incorrect answer. And (as you may have concluded from the fact that I'm posting this at all) I CAN'T FIND MY ERROR!

P.S. This belongs in the suggestion forum, but can a great and holy administrator maybe add an icon? The :banging head against brick wall: icon?

Last edited: Jan 31, 2007
2. Feb 1, 2007

Andrew Mason

Some of the ice (mass = x) will melt but all will be brought to 0 degrees, so:

$$c_im_i(T_f-T_{ice initial}) + L_ix = c_wm_w(T_{water initial}-Tf)$$

$$x = \frac{c_im_i(T_f-T_{ice initial}) - c_wm_w(T_{water initial}-Tf)}{L_i}$$

AM

3. Feb 1, 2007

mbrmbrg

That's true for part a, with two ice cubes. But in part b, the water contains enough energy to melt all of the ice (without, of course, freezing itself).

But I tried entering 0oC as the final temperature, anyways, just to check. WebAssign says no.

4. Feb 1, 2007

mbrmbrg

OK, found one mistake, but still wrong answer. The equation I should be using for part b ought to read

$$-c_wm_w \Delta T = c_im_i \Delta T_1 + L_im_i + c_wm_i \Delta T_2$$

where $$\Delta T_1$$ is the change in temperature from the ice's initial temperature to its melting point, and $$\Delta T_2$$ is the change in temperature from the melted ice to the final temperature of the system at equilibrium.

Shove in numbers to the equation and get

$$(4190)(0.290)(T_f-25) = (2220)(0.045)(0-(-15)) + (333000)(0.045) + (4190)(0.045)(T_f-0)$$

Actually, I'm not quite sure about the specific heat of my ice. I'm using Halliday's Fundementals of Physics 7e, which lists the heat capacity of ice (-10oC) as 2220 J/kg*K. But my ice is -15 degrees. Ice is the only substence in the table which gets a little temperature next to it; does ice's specific heat change as its temperature differs? Well, since I can't find another value (I looked it up at http://www.engineeringtoolbox.com/ice-thermal-properties-d_576.html, but that table didn't agree with Halliday's, so I don't want to use any of its values), I just used 2220 and hoped.

Anyway, do some fun arithmetic with the above equation and beloved calculator says it turns into

$$-1215.1T_f + 30377.5 = 16483.5 + 188.55T_f$$

move T_f's onto one side, divide both sides by T_F's coefficient and get the final temperature = 9.642 oC.

Still the wrong answer, but now I really need a slamming head into brick wall icon.