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Thermodynamics Problems

  1. Jul 1, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    I am working through a couple of revision sheets and have questions about the following two questions:

    1) A)An ice cube of mass 0.03 kg at 0C is added to 0.2 kg of water at 20C in an insulated container.
    (a) Does all the ice melt? (b) What is the final temperature of the drink?
    B) More of a conceptual question: When an object is thermally insulated, that means no heat can enter (escape) into (out of) a system. If this is so, one of the other questions I was doing asked: When a box of strawberries is wrapped in polystyrene (thermally insulating) after coming out of the fridge, how long will it take the strawberries to reach room temperature? My question is: If it is completely thermally insultaing, why must there be a change?

    2)A certain heat source (a heat reservoir or thermal reservoir), is always at temperature ##T_0##. Heat is transferred from it through a slab of thickness ##L## to an object which is initially at a temperature ##T_1 < T_0##. The object, which is otherwise thermally insulated from its surroundings, has a mass m = 0.5 kg and a specific heat capacity ##c = 4 \times 10^{3} J kg^{-1}K^{-1}##. Heat is conducted through the slab at a rate (in J s^{-1}) specified by the
    formula ##KA((T_0 - T)/L),## where K is the thermal conductivity of the slab, A is the area of the
    slab through which heat is transferred and T is the instantaneous temperature of the object.

    a)Show that provided certain assumptions are made, ##KA((T_0 - T)/L)\Delta t = mc \Delta{T}##
    b)Show that for small t, and hence small T, the above equation can be rearranged and
    integrated to give $$T_2 - T_1 = (T_0 - T_1)\left(1 - \exp\left(-\frac{KA(t_2 - t_1)}{Lmc}\right)\right)$$

    3. The attempt at a solution

    1) Both parts are fine - for a), energy required to melt ice cube = E = mL ≈ 10kJ. Energy available (from the water in insulated container) = cmΔT = (0.2 kg) (4.2 kJ kg-1K-1)(293.15K) = 246.3kJ. So the ice will melt. My question is in the so called solution they have available energy as 16.8kJ, i.e have used T = 20 degrees celcius which is incorrect without changing the other units in the equation for the other quantites. However, the answer they get for the final temperature seems more likely than what I attained. With their 16.8kJ, they get a final temperature of ≈ 7 degrees celcius, while I get Tf = (236.2)/((4.2)(0.23)) ≈ -29 degrees celcius. Their answer can't be right because of the unit conversion, however, am I correct?

    2) a)Assuming all energy gained by the slab is then gained by the object ( and there is then no losses because of insulation), $$\frac{KA(T_o - T)}{L} \Delta t = cm (T + \Delta T - T)$$ and result follows.
    b)Integrate $$\frac{KA}{L} \int_{t_1}^{t_2} dt = cm \int_{T_1}^{T_2} \frac{dT}{T_o - T}$$
    which gives $$\frac{KA}{Lmc}(t_2 - t_1) = \ln\left(\frac{T_o - T_1}{T_o -T_2}\right).$$ It is given that ##T_o > T_1## so the numerator of the log is definitely postive. However, I am not so sure about the denominator. Since there are no heat losses, I think there may exist a temperature ##T_2## where ##T_2 >T_o##.

    Many thanks.
     
    Last edited: Jul 1, 2013
  2. jcsd
  3. Jul 1, 2013 #2

    gneill

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    It's preferable to post one question per new thread. I'll address your first question here.
    For the available energy from the liquid water you are interested in a ΔT, not an absolute temperature. The water would freeze if it dropped to the temperature of the ice, so the available ΔT is 20C. Since degrees absolute and degrees Celsius are the same size, ΔT is the same on either scale. So the use of ΔT = 20K = 20C is correct.
     
  4. Jul 1, 2013 #3

    CAF123

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    Ok, yes, my post did look a little cluttered. I just thought each question I had would not require an extensive response so I put them together. At this stage, should I put the others in another thread?

    Okay, thank you, it makes sense. My problem was using an absolute temperature, as you said.
     
  5. Jul 1, 2013 #4

    gneill

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    Perhaps, unless someone jumps in right away to address it here, it'll get more attention as a stand-alone question.
    You're very welcome.
     
  6. Jul 1, 2013 #5

    CAF123

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    I'll put 2) in another thread and maybe keep 1) b) here since it is very short.
    In 1) b), a box of strawberries at initial temperature 25 degrees celcius was put in a fridge and left to cool until it was at the fridge temperature. (Does this mean the heat energy from the strawberries is transferred to the heat energy of the air in the fridge and warms the air up? If so, is this why some fridges have some sort of fan installed in them?)

    The box was then taken out the fridge, wrapped in polystyrene (thermally insulating) and left in the kitchen. How long did it take to get to room temperature?

    The solution gives a finite time however, doesn't thermally insulating imply no heat loss to surroundings? So why would it ever get to room temperature?
     
  7. Jul 1, 2013 #6

    gneill

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    A fridge tries to keep its internal temperature at a constant value by pumping heat out to the world at large (i.e. the kitchen). The internal fan is used to circulate the air inside and move cold air from the freezer compartment to the less-cold area of the general fridge space. It also plays a role in moving hot air (heating elements) across the cooling coils in the freezer on a periodic basis to effect the defrost cycle and prevent ice buildup in the freezer.

    No real insulating material is perfect, so to answer the question you'd need to know something about the thermal properties of the insulating material, its thickness and surface area, the heat capacity of the box contents, and the ambient room temperature.

    If the box was wrapped prior to being placed in the fridge and the time given for the contents to reach the fridge's set temperature, then you might be able to work out the solution from the given information by determining net thermal properties (think Newton's law of cooling).
     
  8. Jul 2, 2013 #7

    CAF123

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    So the question is asking how long it takes the strawberries to reach room temperature, assuming that the wrapping approximates to a thermal insulation?

    If we use the term thermally insulating, in the sense that the wrapping is completely thermally insulating, then the strawberries would never reach room temperature?
     
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