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Unformatted text preview: to solve for n. Because it is a closed system, you can use this n throughout your calculations.) -After your table is filled in, you can draw your PV curve accurately with each state labeled. Waiting until after you are done your table also prevents you from accidentally drawing any processes backwards. -Remember, for an ideal gas, ΔU=nCvΔT whether volume is constant or not. -Also remember to use Cv=3/2R for monoatomic gases and 5/2R for diatomic gases. -In your second table, you might have to do one integral for work. Which sucks if you haven’t learned integrals yet. However, recall the following, and don’t integrate when not necessary! -isochoric, work=0; -isobar, work=p(V2-V1) -adiabat, work: ΔU=-W (use ΔU=nCvΔT) -isotherm, work=∫pdV. Here’s where you have to integrate. Don’t panic! The integral is W = nRT∫dV/V = nRTln(V2/V1). Just memorize that if you have to. -To calculate efficiency, use the positive values of Q from your second table e=Work/Qh. -Also, because with a heat engine you are looking at a cyclic process, Uinitial-Ufinal, ALWAYS. Therefore, Qtotal=Wtotal for the cycle. -When they ask for ΔS, they will invariably ask for one you don’t have to integrate. -isothermal: ΔS=Q/T -Note how all your answers can from the 2 tables you made. -Don’t forget, internal energy is PATH INDEPENDENT. So if there is a question like “How would the internal energy change is process 1 was switched from adiabatic to isochoric?” The answer is NOT AT ALL IF the initial and final conditions are the same. Problem 6 Solution Two cubes of equal side length, one of aluminum and the other of copper, are arranged between heat reservoirs at 100°C and 0°C as shown in diagram a. The heat reservoir at 0°C consists of a mixture of water and ice. The heat conducted by the cubes melts 1250 g of ice every 10 minutes. (a) What is the length of the side of a cube? (b) What is the temperature at the junction between the aluminum and copper cubes? (c) How much ice would melt i...
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This document was uploaded on 02/13/2014.

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