Finding the final temperature of a mixture of ice and steam

  • #1
ayans2495
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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
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Answers and Replies

  • #2
haruspex
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Is the heat relinquished by the steam after condensation positive or negative?
Is Tf-Ti for the steam positive or negative?
 
  • #3
ayans2495
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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.
 
  • #4
Alpher-Bethe-Gamow
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Excellent question, well tackled so far. I would ask you to consider the question; is the change in temperature the same as the difference in temperature between the two states of water?
 
  • #5
ayans2495
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I am not entirely sure what you mean, if your referring to the change in temperature of the steam than I believe that they are equal because the process in which they underwent to change their phase was isothermal.
 
  • #6
Alpher-Bethe-Gamow
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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,
 
  • #7
haruspex
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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.
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.
 
  • #8
haruspex
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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,
It won't. You are making it far more complicated than the question setter would have intended.
 
  • #9
ayans2495
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I got a final temperature of 12 degrees Celsius using my formula, do you have the same?
 
  • #10
Alpher-Bethe-Gamow
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It won't. You are making it far more complicated than the question setter would have intended.
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.
 
  • #11
haruspex
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I got a final temperature of 12 degrees Celsius using my formula, do you have the same?
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.
 
  • #12
Alpher-Bethe-Gamow
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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,
 
  • #13
haruspex
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question setters sometimes ignore air resistance etc. but I don't think they will ignore the laws of thermodynamics
I'm really not sure what point you are making. All I see wrong in post #1 is a sign error.
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.
My interpretation of the equations in post #1 is that @ayans2495 is using Tf and Ti on the steam side (RHS) to mean the final and initial temperatures of the condensed steam respectively, the latter being 100C. So the Tf- Ti will produce a negative value. A positive one is needed.
 
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  • #14
ayans2495
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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.
Would i make the equation as followed:
Q1 + Q2 = -(Q1 + Q2)
I'd really appreciate it if you'd show me your solution, my test is tomorrow though I understand that you're trying to slow walk me to an epiphany.
 
  • #15
Alpher-Bethe-Gamow
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My interpretation of the equations in post #1 is that @ayans2495 is using Tf and Ti on the steam side (RHS) to mean the final and initial temperatures of the condensed steam respectively, the latter being 100C. So the Tf- Ti will produce a negative value. A positive one is needed.
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!
 
  • #16
ayans2495
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I beg of you, pls show me the adjustment mathematically. I'd be eternally grateful. My test is tomorrow.
🥺
 
  • #17
Steve4Physics
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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.
 
  • #18
ayans2495
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I thought so. This is great! Though would I need to do the same for Q3.
 
  • #19
Steve4Physics
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I thought so. This is great! Though would I need to do the same for Q3.
What is the physical meaning of Q3?
Do you think Q3 should be a positive quantity in your equation?
 
  • #20
ayans2495
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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.
 
  • #21
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Keeping the signs correct in this problem using heat flows is certainly not easy. It is much easier to solve this problem by setting the initial internal energy of the system equal to the final internal energy. If we take the internal energy per unit mass of liquid water at 0 C as zero, then the internal energy per unit mass of ice as 0 C, steam at 100 C, and liquid water at temperature T (between 0 C and 100 C) are: $$u_{ice}=-\lambda_m$$
$$u_{steam}=C(100)+\lambda_v$$and $$u_{water}=C(T-0)$$
So $$0.1u_{ice}+0.02u_{steam}=0.12u_{water}$$
 
  • #22
Steve4Physics
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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.
"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.
 
  • #23
Alpher-Bethe-Gamow
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Yes that is right - the Q = mcΔT and Q = mL don't have negatives. Thank you Chestermiller that is a good way to look at it,
 
  • #24
Alpher-Bethe-Gamow
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7
"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.
This came up just as posted, yes perfectly put.
 
  • #25
ayans2495
58
2
"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.
 
  • #26
ayans2495
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2
This came up just as posted, yes perfectly put.
Thank you for your help and input, it really saved me.
 
  • #27
ayans2495
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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.
 
  • #28
ayans2495
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2
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 Tf and Ti on the steam side (RHS) to mean the final and initial temperatures of the condensed steam respectively, the latter being 100C. So the Tf- Ti will produce a negative value. A positive one is needed.
Did you get a final temperature of 42 degrees Celsius.
 
  • #30
ayans2495
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I actually meant to say 41 but I am glad it is close. Thank you so much for taking your time to help me! It really saved me:)
 
  • #31
kuruman
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To @ayans2495:
Something that might save you from future grief is the good habit of using different subscripts for different entities. The default is that the same symbol with the same subscript is the same quantity. In posting #1 you say
miceLf + micecw ΔT= msteamLv + msteamcwΔT
I would use different subscripts to distinguish the two different changes in temperature, and then replace in the next equation ΔTice = Tf - Tice and ΔTsteam = Tf - Tsteam where Tice = 0 oC and Tsteam = 100 oC.
 
  • #32
ayans2495
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2
Noted. Thank you!
 

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