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Problem of Thermodynamics

  • Thread starter ashvuck101
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  • #1
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Homework Statement



An M gram chunk of hot iron is placed on a large piece of 0 ºC ice, causing m grams of ice to melt.

a. What will be the final temperature of the system?
b. What must have been the initial temperature of the hot iron?
c. What must have been the initial temperature of the hot iron?
d. Calculate part c) based on the following information M = 250g, m = 48g and the specific heat of iro in 0.44J / g ºC.

Homework Equations



q=mC(tf-ti)

The Attempt at a Solution



how am i ment to know what the final heat is without being given the heat of the iron all i know is it is above 0 ºC unless i am giving the answer as an equation?

am i missing something really obvious?
 

Answers and Replies

  • #2
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I think they want answers in variable form (equation form) for a,b and c

Then they give you more info to get numerical answers in question d.
 
  • #3
Andrew Mason
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Homework Helper
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Homework Statement



An M gram chunk of hot iron is placed on a large piece of 0 ºC ice, causing m grams of ice to melt.

a. What will be the final temperature of the system?
b. What must have been the initial temperature of the hot iron?
c. What must have been the initial temperature of the hot iron?
d. Calculate part c) based on the following information M = 250g, m = 48g and the specific heat of iro in 0.44J / g ºC.

Homework Equations



q=mC(tf-ti)

The Attempt at a Solution



how am i ment to know what the final heat is without being given the heat of the iron all i know is it is above 0 ºC unless i am giving the answer as an equation?

am i missing something really obvious?
The final temperature is obvious. When the iron stops melting ice, what can you say about the flow of heat from the iron to the ice? What can you then say about the temperature difference between the iron and ice?

For some reason, b and c are the same question.

AM
 
  • #4


The final temperature is obvious. When the iron stops melting ice, what can you say about the flow of heat from the iron to the ice? What can you then say about the temperature difference between the iron and ice?

For some reason, b and c are the same question.

AM
I have exactly the same problem
is it possible to incorporate equation Q=mLf?

c1m1(Tf-Ti)=m2Lf

if so how would this be rearranged? would this be correct

Ti=((m2Lf)/(c1/m1))-Tf

so confused will appreciate any help at all
 
  • #5
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what is Lf?
 
  • #6
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Hi there,

[tex]L_f[/tex] is the latente heat of ice.

The solution of this problem is quite simple. You must apply the energy consrvation law, and you find the answer. It takes heat/energy to transform 0°C ice into water ([tex]\Delta Q=mL_f[/tex]), it takes heat to warm up the water ([tex]\Delta Q=c_{H_2O}m\Delta T [/tex]), and this heat is taken from the hot iron mass ([tex]\Delta Q= c_{Fe}M\Delta T[/tex]). You add all of that, and you find your answer.

Cheers
 
  • #7


thank you, but im still a bit confused

i didnt incorporate [tex]\Delta Q=c_{H_2O}m\Delta T [/tex] because it dosnt say anything about the water changing temperature. only that it melts, so figure it was only a phase change. if the water only changes phase and remains at zero degrees does that not omit that part of the equation. the reason i say this is because the question suggests there is enough ice to absorb enough energy to bring the iron to zero degrees as well. is that right?

ok is this right?
[tex]\Delta Q= c_{Fe}M\Delta T=mL_f+c_{H_2O}m\Delta T [/tex]

rearranged to find initial temp of iron
[tex]\Delta Q= T_{Fe i}=((ml_f+c_{H_2O}m\Delta T)/(c_{fe}m_{fe}))-T_{Fe f}[/tex]

sorry bout the writing im not used to the coding but i hope u get what i mean.
 
  • #8
Andrew Mason
Science Advisor
Homework Helper
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thank you, but im still a bit confused

i didnt incorporate [tex]\Delta Q=c_{H_2O}m\Delta T [/tex] because it dosnt say anything about the water changing temperature. only that it melts, so figure it was only a phase change. if the water only changes phase and remains at zero degrees does that not omit that part of the equation. the reason i say this is because the question suggests there is enough ice to absorb enough energy to bring the iron to zero degrees as well. is that right?
You are correct. The temperature of the water/ice does not change. Some of the ice just melts.
ok is this right?
[tex]\Delta Q= c_{Fe}M\Delta T=mL_f+c_{H_2O}m\Delta T [/tex]
No. I don't think you mean that the two [itex]\Delta T's[/itex] are the same, but that is what you wrote. The first is the change in temperature of the iron and the second is the change in temperature of the water, which, as you have said, is zero.
rearranged to find initial temp of iron
[tex]\Delta Q= T_{Fe i}=((ml_f+c_{H_2O}m\Delta T)/(c_{fe}m_{fe}))-T_{Fe f}[/tex]

sorry bout the writing im not used to the coding but i hope u get what i mean.
The following is not correct:

[tex]\Delta Q= T_{Fe i}[/tex]

Heat flow cannot be equated to temperature.

The following part is correct except for a missing minus sign, but you can simplify because that [itex]\Delta T[/itex] is zero:

[tex]T_{Fe i}=((ml_f+c_{H_2O}m\Delta T)/(c_{fe}m_{fe}))-T_{Fe f}[/tex]

The heat flow equation has to contain the proper signs. Heat flow out is negative and heat flow in is positive. From the perspective of the iron:

[tex]Q_{iron} = - Q_{ice}[/tex]

[tex]m_{iron}c_{iron}(T_f - T_i) = - m_{ice}L_f[/tex]

AM
 
Last edited:
  • #9


thank u so much,
so after well cancel out n simplify is it safe to say

[tex] T_i = (- m_{ice}L_f)/(m_{iron}c_{iron})-T_f [/tex]

n once again guys thankyou for all your help its hugely appreciated.
 
  • #10


although this dosnt seem right when i plug in the given variables in part D
iron M = 250g, ice m = 48g and the specific heat of iro in 0.44J / g ºC.
given Lf ice= 334J/g

Ti iron = (-48x334)/(0.44x250)-0 = -145.7454545 ? that cant be right
 
  • #11


ok i just realised a mistake
-Ti iron = (-5g x 334J)/(0.44J/g x 250g) = 15.18
does that sound right?

thanks
 
  • #12
Andrew Mason
Science Advisor
Homework Helper
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325


although this dosnt seem right when i plug in the given variables in part D
iron M = 250g, ice m = 48g and the specific heat of iro in 0.44J / g ºC.
given Lf ice= 334J/g

Ti iron = (-48x334)/(0.44x250)-0 = -145.7454545 ? that cant be right
Why is this not right? You melt 48 g. of ice requiring 334J/g for a total of 16,032 J. It takes 110 Joules to raise the temperature of the iron one degree C. (250 x .44) J/deg.

AM
 
  • #13


yep sorry confused myself with another question that melts 5g of ice. thanks for all the help guys
 

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