Consider the following example of two substances at different temperatures excha

Question

Consider the following example of two substances at different temperatures exchanging heat as they come to thermal equilibrium. A 3.50 kg piece of lead and a 1.5 kg piece of aluminum, at different temperatures, are placed in contact. They are able to exchange energy as heat, and they can only exchange it with each other. The initial temperature of the lead is 48'C, and the initial temperature of the aluminum is 35degreesC. The table on page 5 of your textbook has phase change temperatures and values of specific heats for these substances. If it is possible, construct a single, closed energy-system diagram for the process of lead and aluminum coming to thermal equilibrium. How do you know this? Write an algebraic expression of energy conservation in terms of the AEs of the two energy systems, and then substitute in the appropriate expressions for the AEs. Finally, substitute in all known numerical values. How many unknowns are there in your equation expressing energy conservation? At an intermediate point in the entire process described above the lead has reached a temperature of 43degreesC. Follow the instructions for part (a) for the interval that ends when the lead is at a temperature of 43degreesC. What feature related to the process connects the two final temperatures that appear in your diagram to each other?

Explanation / Answer

m = 3.5 kg lead

m = 1.5 kg aluminum

Tlead i = 48°C

Tal i = 35°C

Cp lead = 0.13 kJ/kgC

Cp alu = 0.91 kJ/kgC

a)

Qaluminium = -Qlead

Qaluminium = mal*Cpal*(Tf-Tal i)

Qlead = mlead*Cplead*(Tf-Tlead i)

mal*Cpal*(Tf-Tal i) =-mlead*Cplead*(Tf-Tlead i)

NOTE that Tf in both cases is the same (thermal equilibrium)

I know this because its the first law of thermodynamics, that is, energy is not created nor destroyed. Here we are exchangin heat

Number of unown (Tf, 1 eqn, you can solve it)

-mal*Cpal*(Tf-Tal i) =mlead*Cplead*(Tf-Tlead i)

1.5*(0.91)(Tf-35) = (3.4)*(0.13)*(Tf-48)

Solving for Tf

1.365Tf - 47.77 = 0.442Tf -21.22

0.925Tf = 26.55

Tf = 28.7°C

b)

When T lead is at 43, nothing is changed, you can still use the formula since th lead is transfering heat to aluminum

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