A kitchen oven has an interior volume V of 0.12 m^3. The air in the oven is preh

Question

A kitchen oven has an interior volume V of 0.12 m^3. The air in the oven is preheated from an initial temperature T0 of 20C to a baking temperature T1 of 175C. The initial pressure p0 inside the oven equals the outside pressure of 1000hPa.

a) If the oven is sealed (i.e., the process is isochoric), how much heat is added to the air in the oven?

b)If the oven is not sealed (the normal case), air will gradually leak out as the air expands from heating, so that the process is isobaric. Assume that the air leakage is one-way (no outside air enters the oven). How much total heat is added to the air? (Hint: You are heating not only the air that remains in the oven at the end of the heating process but also some of the air that escapes. You will have to set up and integrate a differential equation. Assume that the air temperature inside the oven is always uniform at any point in the process and equals the temperature of any air leaking out.)

Explanation / Answer

(a) initially p V = n R T or nR = p V / T = 100000 * 0.12 / 293 = 40.956

Also Q = deltaU = (5/2) n R deltaT because work is zero (Q is heat added)

So... Q = (5/2) * 40.956 * (175-20) = 15870 Joules

(b) Now for constant pressure...

Split the air into two parts: the part that remains in the oven and the part that leaves the oven. For the part that remains in the oven, it's pressure and volume remains constant. Which means since pV is constant, nRT remains the same for the air that is in the oven. This means that the total internal energy, U, remains the same for the air in the oven. And since no work was done... the net heat added to the air that remains in the oven is zero!! This is weird, but true. Welcome to thermodynamics.

This also means that the net heat added to the air is only the heat added to the air that leaves the oven. Imagine that:

we increase the temp of the oven in small steps, dT

when we do this, the volume of the air inside the oven expands... dV

and the number of moles of air in this volume, dn, leaves the oven.

When that air leaves the oven, it takes with it the internal energy of the air which is dU = (5/2) RT dn

If we integrate this, to find all the internal energy the air takes out of the oven, we get

U = integral of (5/2) R (pV/nR) dn = (5/2) pV integral of (dn/n) =

= (5/2) pV ln(n final / n initial) =

= (5/2) pV ln ( T initial / T final) =

= (5/2) 100000 * 0.12 * ln (293 / 448) = -12739 Joules

This means 12739 Joules left the oven with the internal energy of the air that leaked out.

But that's not all... there was also work done to push that air out.

Work done to expand the air dW = p dV

dW = n R dT

= (pV / T) dT   

= pV ( dT/T )

Work = pV integral of ( dT/T ) =

= pV ln(T final / T initial) =

= 10000 * 0.12 * ln(448/293) = 5095 Joules

So the total heat added to the air is 12739 + 5095 = 17834 Joules

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