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Heat Transfer by Nichrome wire and Steel

  1. Sep 29, 2015 #1
    Hello all. I want to do some experiment but I thought I would go over this theoretically before trying out and now I have a doubt.

    1. The problem statement, all variables and given/known data
    1. I have a driver which produces say 600mA. I connect the driver to one end of Nichrome Wire. And the other end of Nichrome wire is in contact with A Mild Steel substance.
    2. I want to know (my main doubts are) (a)-How much heat is generated in the Nichrome Wire due to the current in 1 second? (b) For 1 second, what will be the change in temperature at the mild steel due to the current produced by driver? (Assuming initial temperature= the room temperature .i.e 24C).
    Now, I did a lot of research and came up with this. I want to know know whether I am heading in a right path. If not, I need help with this. I will simplify what I have done till now.

    2. Relevant equations
    • Joules law= H=(I^2)*R*t
    • Specific Heat Capacity= Q=cp*m*dT
    • time required to change temperature of object=t=(m*cp*dT)/H
    3. The attempt at a solution
    1. By Joules's law of heating, H=(I^2)*R * t. where I=600 mA, R of Nichrome wire is 2.196 ohms. Therefore heat emitted by Nichrome wire is H=0.79J for 1 second. DOUBTS - Is 0.79J of heat is there in the Nichrome wire in 1 second. And does it mean the heat present in Nichrome wire for 2 seconds is 0.79*2 ?

    2. Temperature change in Nichrome wire dT= (h*t)/(Cp*m); where H=0.79J(from above answer);t=1 sec;Cp=specific heat capacity of nichrome=0.45 J/gC; mass=10 g. After evaluating, I get dT=0.175 C. So, for 1 second, temperature of Nichrome rises by 0.175 C ?

    3. Now, I need temperature change in mild steel for one second. So, I used this formula, Time required to change temperature of mild steel=(m*Cp*dt)/H. H=0.79J (from point-1) Here, mass of mild steel=5 g; dt=0.175 C (from above answer); Specific heat capacity of mild steel= 0.51 J/gC. After evaluating, I get t=0.56 seconds. This means, for 0.5 seconds, temperature in steel changes by amount of 0.175 C. So, I need 3.36 seconds for 1 C change.
    Note- I know that there will be many factors under consideration. For example, There will be loss of heat to environment. And resistance of Ni-wire change over temperature. But, can we try to predict the temperature change in mild steel assuming there is no loss and no changes in wire over time?

    I know this is a big question and difficult to answer here. But any small help is appreciated as I am confused with this.
     
    Last edited: Sep 29, 2015
  2. jcsd
  3. Sep 29, 2015 #2

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    Yes.
    Yes.
    Heat capacities are additive. Recalculate "3."
     
  4. Sep 29, 2015 #3
    Thanks for the reply.
    Ok, so only 3rd part is wrong? I redid the calculations.
    • I used the formula, q=m*c*dT. where m=5g, c=0.51 J/gC. dT=0.175C. I got q=0.446J. So, I need 0.446J to raise temperature of steel by 0.175 C. But I have 0.79J(from answer 1) available. By cross multiplication, the evaluation comes to 0.3C. So, I can raise the temperature of steel by 0.3 C for the incoming heat of 0.79J per second.
    Is am sensing even this is wrong as this answer is similar to the previous one. Can you help me out with this one?
     
  5. Sep 29, 2015 #4

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    Additive.
     
  6. Sep 29, 2015 #5
    I honestly can't think of any other way to do it. Should I change the formula or modify the answer with the same formula? Logically thinking, I think the formula is right but I have to add some other quantity to it? Sorry for bothering you again.
     
  7. Sep 29, 2015 #6

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    In parts 1) & 2) you were heating nichrome alone. What are you heating in part 3)?
     
  8. Sep 29, 2015 #7
    Ok. Here, I am heating Steel. Need some clarification. Again got confused.
    1. Heat will transfer from hot body to cold body till the cold body temperature=hot body temperature right? Now after 1 second, Nichrome will be at 0.175 C ( Assume initial temperature = 0 C for calculation convenience).
    2. So, the hot body is Nichrome which is at 0.175 C and cold body is steel which is at 0 C. (Let us assume the transfer of heat starts after 1 second even though practically it will start before 1 second).
    3. Now, heat will transfer till both the particles in contact are at same temperature. So, some amount of heat ( equivalent to 0.175 C) will be transferred to Steel in 0.56 seconds( from point 3 in main post ). So, at 1.56 seconds ( 1 second for Nichrome, 0.56 seconds for Steel), both particles are at 0.175 C.****** Now, won't there be any temperature change in both the particles as both are equally hot? ( But theoretically, temperature should increase in both as Heat is being accumulated at every instant ) If we do calculation like this, Steel will get to 1 C only after Nichrome goes to 1 C. i.e, at 6th second. Right?
    • ****** -> But at 1.56 seconds, some amount of heat is added to the Nichrome from the Driver. So, it's temperature must have increased by 0.075 C(0.175/2) making it's total of 0.2625 after 1.5 seconds. So, now, again heat transfer takes place as both bodies are not at same temperature.
    I am sorry. I think I have confused you with my questions.
    Here is what I thought might happen...

    TIME(s) --------------- NICHROME (C)--------------- STEEL (C)
    0 ------------------------ 0 ------------------------------- 0
    1 ----------------------- 0.175 -------------------------- 0
    2 ----------------------- 0.3 -------------------------------- 0.3125
    3 ----------------------- 0.525 --------------------------- 0.6125
    4 ----------------------- 0.7 -------------------------------- 0.9375
    5 ----------------------- 0.875 --------------------------- 1.25
    6 ----------------------- 1.05 -----------------
     
    Last edited: Sep 29, 2015
  9. Sep 30, 2015 #8

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    Nichrome AND steel. Add the heat capacities.
     
  10. Sep 30, 2015 #9
    Ok. Hopefully this is right.

    • t=(m*cp*dT)/H; where m is mass of steel=5; dT= 0.175 C; H=0.79 J; cp= 0.51+0.45=0.96 g J/g C
    • So, t=(5*0.96*0.175)/0.79= 1.05s.
    • To increase temperature by 0.175 C, steel takes 1 second. So, time taken by steel to raise temperature of 1 C=~6 seconds.

    Is this also wrong?
     
  11. Sep 30, 2015 #10

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    How much nichrome? How much steel? Why are you multiplying mass of steel by the sum of the two specific heats?
     
  12. Sep 30, 2015 #11
    You told me to add heat capacities. I mistook it as specific heat and added that and used it in the formula.
    • So, Heat capacity formula is c=ΔQ/Δ T. where Δ Q is the amount of heat transferred, Δ T is the temperature difference. But, I can't add heat capacity of steel and nichrome because q is same for both and ΔT is common between thosetwo.
    • Specific heat capacity of nichrome is 0.45 J/g C and of steel is 0.51J/g C. If I add both I will get 0.91 J/g C.
    • Here, I have to take in account of both nichrome and steel because Heat is passing through the nichrome to get to steel. So, I have to consider nichrome too to get the value of steel. I can understand that. But I don't know what to add ?
    • Can you tell me which formula to use and why to use that ?
    • And what do you mean by- How much nichrome? How much steel?
    • Or should I use this formula- Q = kA(THot−TCold)/d ?
    I am sorry. I know, I am dragging this but I simply can't understand the last part.
     
  13. Sep 30, 2015 #12

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    What is the heat capacity (not specific heat) of 10 g of nichrome? What is the heat capacity of 5 g of steel? What is the sum of those two heat capacities?
     
  14. Sep 30, 2015 #13
    • Heat Capacity= mass * specific heat
    • Nichrome- C=10 * 0.45= 4.5 J/C
    • Steel- C=5 * 0.51= 2.55 J/C
    • So, I will get 7.05 J/C ?
     
  15. Sep 30, 2015 #14

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  16. Sep 30, 2015 #15
    • Now, I have to use it here- ? time= (C*dT)/H where C= 7.05J/C, dT=0.175 C, H=0.79 J. So, I will get time=1.56 seconds.
    • So, 1.56 seconds needed for steel to increase temperature by 0.175 C. So, 9.36 seconds to increase temperature of steel by 1 C ?
     
  17. Oct 1, 2015 #16

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    No.

    You are mixing up the concepts of "power" (energy per unit time) and energy. Check Wiki, or some other source, and get those two ideas sorted out, then get back to this thread.
     
  18. Oct 1, 2015 #17
    I looked them up. My last attempt. Hopefully it's right.
    • Power is rate of consuming Energy. (W)
    • Energy tells quantity of work done. (J)
    • Heat Capacity tells amount of heat required to change temperature of whole system by 1 C. Total heat capacity here is 7.05 J/C.
    • But the heat available per second from Nichrome wire is 0.79J. Heat has to pass through Nichrome to get to steel. So, I need in total of 8.92 seconds to raise temperature by 1 C ?
     
  19. Oct 1, 2015 #18

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    Close enough.
     
  20. Oct 1, 2015 #19
    Oh ok. Thanks a lot. My last doubt ( An overview of this ) - more like a yes/no question.
    • I supply 500mA of current from driver to Nichrome wire. So, because of this, heat generated in Nichrome wire is 0.79 J in 1 second. So, in 2 second 1.58 J.
    • This heat is transferred to the Steel every second. ( every second means, practically, when Nichrome's temperature rises even a little bit more than steel's)
    • So, in close to 9 seconds, the temperature of Steel rises by 1 C. And, in 18 seconds since the start period, temperature of Steel rises by 2 C and so on...
     
  21. Oct 1, 2015 #20

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