A processor chip sports a big heatsink into which air a diat

A processor chip sports a big heatsink into which air (a diatomic gas) at room temperature (30.3 oC) is blown by a fan at speed 6.1 m/s. The fan/heatsink combination has a orfice of about 7.2 cm (a) If the processor is consuming 127.2 W, what is the temperature increase in the blown air?

Solution

The first process is an isochoric one, since the volume remains constant

Hence P_a/T_a = nR/V = P_b/T_b

Which gives T_a = P_aV/nR = 4.8K

Now, T_b = PbTa/Pa = 10.68K

U =3/2nRT = 74.8 Joules of energy is exchanged during process 1

b) The process 3 being an isobaric process, work done

W= pV = 19.46(4.05-2.05)=38.92 J of work is done

c) The process 2 is an isothermal process

Work done , w= nRTln Vb/Va = 60.42 J of work is done. Since there is no change in the internal energy, the First Law says that the heat exchanged is equal to the work done.  So, for this process the heat exchanged is also 60.42J

d) Efficiency = 1- T₁/T₂ = 0.55