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mbrmbrg
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Intro Thermodynamics--put ice in water
(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?
Q_out=Q_in
For temperature change without phase change:
[tex]Q=cm\Delta T[/tex]
For phase change:
[tex]Q=Lm[/tex]
oC + 273.15 = K
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) [tex]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 [/tex]
I had found in part a that the energy that the water can give off before freezing is [tex]c_wm_w \Delta T = (4190 \frac{J}{kg K})(0.290 kg)(298.15) = 30377.5 J[/tex].
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:
[tex]c_im_i(T_f-T_{ice initial}) + L_im_i = -c_wm_w(T_f-T_{water initial})[/tex]
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:
[tex](2220)(0.045)(T_f - (-15)) + (333000)(0.045) = -(4190)(0.290)(T_f - 25)[/tex]
This arithmetics out to [tex]99.9T_f + 1498.5 = -1215.5T_f + 30377.5[/tex]
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?
Homework Statement
(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?
Homework Equations
Q_out=Q_in
For temperature change without phase change:
[tex]Q=cm\Delta T[/tex]
For phase change:
[tex]Q=Lm[/tex]
oC + 273.15 = K
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) [tex]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 [/tex]
I had found in part a that the energy that the water can give off before freezing is [tex]c_wm_w \Delta T = (4190 \frac{J}{kg K})(0.290 kg)(298.15) = 30377.5 J[/tex].
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:
[tex]c_im_i(T_f-T_{ice initial}) + L_im_i = -c_wm_w(T_f-T_{water initial})[/tex]
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:
[tex](2220)(0.045)(T_f - (-15)) + (333000)(0.045) = -(4190)(0.290)(T_f - 25)[/tex]
This arithmetics out to [tex]99.9T_f + 1498.5 = -1215.5T_f + 30377.5[/tex]
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?
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