Aclosed, rigid tank fitted with a paddle wheel contains 2.0 kg of air, initially at 200oc, 1 bar. during an interval of 10 minutes, the paddle wheel transfers energy to the air at a rate of 1 kw. during this time interval, the air also receives energy by heat transfer at a rate of 0.5 kw. these are the only energy transfers. assume the ideal gas model for the air, and no overall changes in kinetic or potential energy. do not assume specific heats are constant. determine the change in specific internal energy for the air, in kj/kg, and the final temperature of the air, in oc.

T=833.8 °C

Explanation:

Given that

m= 2 kg

T₁=200 °C

time ,t= 10 min = 600 s

Work input = 1 KW

Work input = 1 x 600 KJ=600 KJ

Heat input = 0.5 KW

Q= 05 x 600 = 300 KJ

Gas is ideal gas.

We know that for ideal gas internal energy change given as

ΔU= m Cv ΔT

For air Cv= 0.71 KJ/kgK

From first law of thermodynamics

Q  = ΔU +W

Heat input taken as positive and work in put taken as negative.

300 KJ = - 600 KJ + ΔU

ΔU = 900 KJ

ΔU= m Cv ΔT

900 KJ = 2 x 0.71 x (T- 200 )

T=833.8 °C

So the final temperature is T=833.8 °C

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