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Thermodynamics - heating spoon and coffee

  1. Nov 20, 2004 #1
    can someone please tell me which equation i need to use for this problem?
    A 50g metal spoon is placed in a cup with 200g of hot coffee. If the spoon's initial temperature is [tex]20\circ C[/tex] and the coffee's temperature is [tex]100\circ C[/tex] , what is the temperature of the spoon and coffee when their temperatures are equal?
  2. jcsd
  3. Nov 21, 2004 #2
    [tex]\mbox{heat gained}+\mbox{heat lost}=0[/tex]
    [tex]Q=mc\Delta T[/tex]
    where Q is heat and c is specific heat capacity. Figure out what loses and what gains heat, substitute the second equation into the first and solve. (Remember lost heat has a negative sign).
  4. Nov 21, 2004 #3
    so is it (50)(c)(80) + (200)(c)(-80) = 0 ?
    do i need to convert temperature to kelvin?
  5. Nov 21, 2004 #4


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    you don't need to change temp to kelvin, and Where did you get 80 for Delta'T'? That is what you are solving for, 'c' is a number that you get from a book or the problem. So plug the values for 'c' in, and solve for Delta'T'.

    (someone needs to confirm this though, i haven't done thermodynamics in a while, i'm a little rusty)
  6. Nov 21, 2004 #5
    i did 100 - 20 = 80 but i guess thats wrong. is c a constant value? i can't seem to find it in the text book
    Last edited: Nov 21, 2004
  7. Nov 21, 2004 #6


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    no, that's not right. C is the specific heat capacity of an object (if i remember correctly, its the # of joules that it takes to raise 1g of water 1 degree celsius, or something like that). The problem should give you the value for each item. 'C' is a different value for each object. So you will have different 'c' values for the spoon and the coffee.
  8. Nov 21, 2004 #7
    the value of c isn't given in the problem. so how would i solve for it?
  9. Nov 21, 2004 #8


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    You can't solve for it.. =/

    afaik its determined experimentally...
  10. Nov 21, 2004 #9
    oh so if its not given in the problem how would i solve for [tex]\Delta T[/tex] ?
  11. Nov 21, 2004 #10


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    To solve for Delta'T' you have to have 'c'...
  12. Nov 21, 2004 #11
    You definitely need the specific heat capacities of the coffee and metal (coffee's shouldn't be much different from 4.18J/g-K) because I see two equations and 3 unknowns so at best you can express two as a relation to the third.
  13. Nov 21, 2004 #12


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    Staff: Mentor

    Hint on [tex]\Delta T[/tex]

    The colder spoon will experience a [tex]\Delta T[/tex] = T-20 °C. The water will experience a [tex]\Delta T[/tex] = 100 °C - T. T is the final common temperature.

    Heat capacity and specific heat are discussed at -

    The change in heat content [tex]\Delta Q[/tex] of an object is just [tex] m \times c_p \times \Delta T[/tex], where [tex]c_p[/tex] is the specific heat and [tex]m[/tex] is mass.

    The specific heat of water is already given, but there are range of specific heats depending on the metal.

    Al ~ 0.9 J/g-K
    Cu ~ 0.39 J/g-K
    Silver ~ 0.23 J/g-K
    Fe ~ 0.45 J/g-K, another site suggests ~0.486 (http://members.shaw.ca/diesel-duck/library/exam_stuff/3rd_app_heat.htm ) similar problem to above but it involves hot ball bearings in cold water - :smile:

    http://web.mit.edu/3.091/www/pt/pert13.html ([tex]c_p[/tex] at 300 K for the elements

    For a theoretical discussion see -

    Here's are practical heat capacity/transfer issue - :biggrin: - http://www.gunownersalliance.com/Rabbi_0168.htm
    Last edited: Nov 21, 2004
  14. Nov 21, 2004 #13
    so assuming that the specific heat capacity of coffee is 4.186 J/g°C and for the spoon its 90.2 J/g°C would the equation be
    (50g)(90.2 J/g°C)(T - 20°C) + (200g)(4.186 J/g°C)(100°C - T) = 0 ?

    T = 18.5°C
  15. Nov 21, 2004 #14
    I'm not sure the [tex]\Delta t[/tex] you were given was right. From the question we know that the spoon will increase in temperature while the coffee will decrease in temperature. We know that their temperatures will be equal at some point. Let the temperature that they are both equal at be x:

    [tex]20+\Delta T_s = x[/tex]
    [tex]100+\Delta T_c = x[/tex]

    Where s and c refer to the spoon and coffee, respectively. Setting these two equal yields:

    [tex]20+\Delta T_s = 100+\Delta T_c[/tex]
    [tex]\Delta T_s = 80 + \Delta T_c[/tex]
    [tex]\Delta T_c = \Delta T_s - 80[/tex]

    We know that the total energy gained by the spoon equals to total energy lost by the coffee so...

    [tex]50\times 90.2 \times \Delta T_s + 200 \times 4.186 \times \Delta T_c = 0[/tex]
    [tex]50\times 90.2 \times \Delta T_s + 200 \times 4.186 \times (\Delta T_s - 80) = 0[/tex]

    You have one unknown and an equation. Solve for it and plug it back into another equation and solve for x.
    Last edited by a moderator: Nov 21, 2004
  16. Nov 21, 2004 #15


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    Staff: Mentor

    T must be between 20° and 100°, assuming that no heat is lost from the system, i.e. adiabatic boundary condtions.

    Try (50g)(90.2 J/g°C)(T - 20°C) = (200g)(4.186 J/g°C)(100°C - T), or

    (50g)(90.2 J/g°C)(T - 20°C) - (200g)(4.186 J/g°C)(100°C - T) = 0

    The spoon gains heat and the water loses heat, the sum (balance) must = 0.

    [tex]Q_s + Q_w = 0 (adiabatic) [/tex], where Q > zero for heat gained and Q < 0 for heat lost, or [tex]Q = m\,c_p\,T-T_o [/tex], where [tex]T and T_o [/tex] are final and initial temperatures. Adiabatic means no loss of heat from the system, so the net gain of heat is zero. It's like the net force (sum of all forces) is zero in a statics problem.
    Last edited: Nov 21, 2004
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