 #1
ayans2495
 58
 2
 Homework Statement:
 100 g of ice at 0 ºC is added to an insulated chamber containing 20 g of steam at 100 ºC. What is the final temperature of the 120 g of water?
 Relevant Equations:

Q = mcΔT
Q = ml
The condensed steam loses heat, so the heat it loses is a positive quantity. This is in addition to the heat it yielded in condensing.That is exactly question. I would assume that it is negative though i didn't get the right answer with it. Maybe my math was wrong. On the contrary, as heat is extracted from the steam, I believe ΔT is negative.
It won't. You are making it far more complicated than the question setter would have intended.Some energy will be given to break bonds and forming bonds releases energy so that this is not an isothermal process, as the temperature in the closed system will change. If there were no changes in states of matter then you would be right, as the temperature change by one would directly be affected by the temperature change in the other. I hope this helps,
I know question setters sometimes ignore air resistance etc. but I don't think they will ignore the laws of thermodynamics :) Also the change of temperature is irrespective of direction in this equation, should point that out as it seems you want to look at that in your posts, 100 > 50 is the same as 0 > 50, no negatives used.It won't. You are making it far more complicated than the question setter would have intended.
Using which formula? The equation in post #1 has a sign error, as I indicated.I got a final temperature of 12 degrees Celsius using my formula, do you have the same?
It should be higher than 12, and haruspex is correct it should be a middling type of number,I got a final temperature of 12 degrees Celsius using my formula, do you have the same?
I'm really not sure what point you are making. All I see wrong in post #1 is a sign error.question setters sometimes ignore air resistance etc. but I don't think they will ignore the laws of thermodynamics
My interpretation of the equations in post #1 is that @ayans2495 is using T_{f} and T_{i} on the steam side (RHS) to mean the final and initial temperatures of the condensed steam respectively, the latter being 100C. So the T_{f} T_{i} will produce a negative value. A positive one is needed.Also the change of temperature is irrespective of direction in this equation, should point that out as it seems you want to look at that in your posts, 100 > 50 is the same as 0 > 50, no negatives used.
Would i make the equation as followed:Using which formula? The equation in post #1 has a sign error, as I indicated.
Please post the equation you are now using and the subsequent working.
I get quite a bit more than 12C.
To get a feel for what the answer ought to be, note that although there is 5 times the mass of ice the latent heat of vaporisation is about seven times that of fusion, so it should be a middling sort of number.
You are right to assume this, and right in he needs to make it all positive, a simple adjustment, and he will have the answer!My interpretation of the equations in post #1 is that @ayans2495 is using T_{f} and T_{i} on the steam side (RHS) to mean the final and initial temperatures of the condensed steam respectively, the latter being 100C. So the T_{f} T_{i} will produce a negative value. A positive one is needed.
What is the physical meaning of Q3?I thought so. This is great! Though would I need to do the same for Q3.
"Q3 is the amount of heat required to condensate the steam"Q3 is the amount of heat required to condensate the steam, as it is releasing heat one would think it is negative. On the contrary, there are no negative values in the equation Q=mlv, so I would suppose that it should be positive.
This came up just as posted, yes perfectly put."Q3 is the amount of heat required to condensate the steam"
No! No energy is required (i.e. you don't need to supply energy).
Have you ever boiled water to make a cup of tea? You *require* (must supply) an amount of energy ##ml_v## to convert water to steam.
But you do not need to supply any energy to condense the steam to water. In fact the steam *releases* an amount of energy ##ml_v## when it condenses.
In words, the equation ##Q_1 + Q_2 = Q_3 + Q_4## means:
Het required to melt ice
+ Heat required to raise ice from ##0^oC## to ##T_f##
=
Heat released when steam condenses
+
Heat released by condensed steam cooling from ##100^oC## to ##T_f##
Expressed this way, ##Q_1,Q_2, Q_3## and ##Q_4## must all be positive quanties.
I understood all that, thank you. I understand that to reverse the process energy is actually extracted from the steam, my English was just poor there. Again, thank you."Q3 is the amount of heat required to condensate the steam"
No! No energy is required (i.e. you don't need to supply energy).
Have you ever boiled water to make a cup of tea? You *require* (must supply) an amount of energy ##ml_v## to convert water to steam.
But you do not need to supply any energy to condense the steam to water. In fact the steam *releases* an amount of energy ##ml_v## when it condenses.
In words, the equation ##Q_1 + Q_2 = Q_3 + Q_4## means:
Het required to melt ice
+ Heat required to raise ice from ##0^oC## to ##T_f##
=
Heat released when steam condenses
+
Heat released by condensed steam cooling from ##100^oC## to ##T_f##
Expressed this way, ##Q_1,Q_2, Q_3## and ##Q_4## must all be positive quanties.
Thank you for your help and input, it really saved me.This came up just as posted, yes perfectly put.
Did you get a final temperature of 42 degrees Celsius.Your original equation
##Q_1 + Q_2 = Q_3 + Q_4##
is based on ##Q_1, Q_2, Q_3## and ##Q_4## all being positive quantities.
##Q_4## is the thermal energy released when the water from the (condensed) steam cools from ##T_i = 100^oC## to ##T_f##.
You have used ##Q_4 = m_{steam} c_w (T_f  T_i)##
But ##T_i > T_f##, so ##T_f  T_i## will be negative and ##Q_4## will be negative, which is wrong.
So you need to make a 'manual adjustment' to allow for this, therefore use
##Q_4 = m_{steam} c_w (T_i – T_f)##
or
##Q_4 = m_{steam} c_w (T_f – T_i)##
which then makes ##Q_4##, positive, as required.
Did you get a final temperature of 42 degrees Celsius.I'm really not sure what point you are making. All I see wrong in post #1 is a sign error.
My interpretation of the equations in post #1 is that @ayans2495 is using T_{f} and T_{i} on the steam side (RHS) to mean the final and initial temperatures of the condensed steam respectively, the latter being 100C. So the T_{f} T_{i} will produce a negative value. A positive one is needed.
Yes.Did you get a final temperature of 42 degrees Celsius.