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CAF123
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Homework Statement
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 $$T_2 - T_1 = (T_0 - T_1)\left(1 - \exp\left(-\frac{KA(t_2 - t_1)}{Lmc}\right)\right)$$
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.
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