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Thermodynamics - hot item placed in water

  1. Sep 9, 2008 #1
    1. The problem statement, all variables and given/known data

    A 900 g aluminum pan is removed from the stove and plunged into a sink filled with 12.0 L of water at 19.0 degrees. The water temperature quickly rises to 24.5 degrees.

    What was the initial temperature of the pan in degrees C?

    What was the initial temperature of the pan in degrees F?

    2. Relevant equations

    Q_net = 0 where Q is heat

    No phase changes involved so no latent heats required

    C1 = specific heat capacity, aluminum: 900 J/kg-K
    C2 = Specific heat capacity, water: 4.184 kJ/kg-K = 4184 J

    Q_net = (C1)(M1)(T) + (C2)(M2)(T2) = 0

    where C1 = 900, C2 = 4184, M1 = 900g = 0.9 kg, M2 = 12kg because 1 L water = 1 kg, T = unknown, T2 = 24 - 19 = 5 degrees

    Tc = T - 273.15 where Tc is temp in celsius, T is temp in kelvin
    Tf = 9/5(Tc) + 32 where Tf is temp in fahrenheit, Tc is temp in celsius

    3. The attempt at a solution

    Q_net = (C1)(M1)(T) + (C2)(M2)(T2) = 0

    no phase changes involved, no latent heats needed

    = (900)(0.9)(T) + (4184)(12)(24-19)
    = (900)(0.9)(T) + (4184)(12)(5)
    T = -[(4184)(12)(5)] / [(900)(0.9)] = -309 degrees

    I assumed that -309 was in kelvin, thus used the above equation to solve for temperature in celsius.

    i noticed that the temp was negative, thus decided that since the water temp increased, the temperature had to be positive, thus -309 became 309, and using the kelvin-->celsius conversion i got 36 degrees Celsius which was incorrect?

    as for the second part, the initial temp in fahrenheit, i need to solve the first part to get it.

    what did i do wrong? it seems like a straightforward problem, i think i missed something.

    thanks
     
  2. jcsd
  3. Sep 10, 2008 #2

    alphysicist

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    Homework Helper

    Hi portofino,

    I think you are misinterpreting what T is. Looking at these steps:

    If you look at the water term, 5 is not the temperature in Kelvin of the water, it is the change in the water's temperature. The T in this equation plays the same role for the aluminum.
     
  4. Sep 10, 2008 #3
    so should T actually be (Tf - Ti) where Tf is final temp and Ti is the initial temp? or should T just become deltaT = change in temp?

    should i even have two unknowns for temperature?

    actually it should have been 24.5 - 19 = 5.5, not 5. how do i use the temperature change (5.5)? am i supposed to find a temperature equivalent for 24.5 and 19? if so i don't know what scale the temperatures are in.
     
  5. Sep 10, 2008 #4
    I don't think this is a very well worded problem. You need to make the assumption that both the water and the pan have reached thermal equilibrium. If that is the case then the pan would have the same final temperature as the water. Meaning that you only have one unknown (temp of pan).
     
  6. Sep 10, 2008 #5
    i see what you're saying topher, so T should then be (24.5 - Ti)?
     
  7. Sep 10, 2008 #6
  8. Sep 10, 2008 #7
    is that the same case for the water, so instead of 5.5 it should also be T = (24.5 - Ti), what happens to the 19 degrees stated in the problem?
     
  9. Sep 10, 2008 #8
    That shouldn't make a difference. 24.5 - Ti = 5.5.
     
  10. Sep 10, 2008 #9

    stewartcs

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    Science Advisor

    Perhaps it would be clearer to re-write your energy balance:

    [tex]m_{water} \cdot C_{water} \cdot (T_f - T_i)_{water}[/tex] + [tex]m_{AL} \cdot C_{AL} \cdot (T_f - T_i)_{AL} = 0[/tex]

    Since they are in equilibrium, [tex] T_{f,water} = T_{f,AL} = T_f [/tex]

    Plug that back into the equation above and the only unknown is [tex] T_{i,AL} [/tex], so it would then be a matter of rearranging the equation and plugging in the given/known values.

    The initial and final temperatures were given in the problem in Celsius. Convert those to Kelvin before solving the equation (i.e. just add 273 to the given values). Also for future reference you may want to remember that Kelvin is an absolute scale. Hence, there can be no negative values in Kelvin.

    Hope this helps.

    CS
     
  11. Sep 10, 2008 #10
    using the above equation i got this:

    Tf_w = Tf_al = 24.5 = 24.5 + 273 = 297.5 kelvin
    Ti_w = 19 = 19 + 273 = 292 kelvin

    find Ti_al

    [(12)(4184)(297.5-292)]_water + [(0.9)(900)(297.5 -Ti)]_aluminum = 0
    271644 + 240975 - 810Ti = 0
    810Ti = 517119
    Ti = 638.41 kelvin, so for part one Ti in celsius = 638.41 - 273 = 365.418 celsius

    and for part b using equation:

    Tf = 9/5(Tc) + 32 where Tf is temp in fahrenheit, Tc is temp in celsius

    temp in fahrenheit = 9/5(365.418) + 32 = 689.753 deg fahrenheit

    is that correct now?
     
  12. Sep 10, 2008 #11
    sorry to double post, just wanted to let you all know that the above values i got were in fact correct.

    thanks for the help
     
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