Question

A rigid tank with a volume of 0.5 m3 contains air at 120 kPa and 300 K. Find the final temperature after 20 kJ of heat is added to the air using (a) constant specific heats and (b) ideal gas tables.

Answer 1

T=340. 47 K

Step-by-step explanation:

Given that

Volume of tank =0.5
`m^3`

Pressure P=120 KPa

Temperature T=300 K

Added heat ,Q= 20 KJ

Given that air is treated as ideal gas and specific heat is constant.

Here tank is rigid so we can say that it is constant volume system.

We know that specific heat at constant volume for air

`C_v=0.71\ (KJ)/(kg.K)`

We know that for ideal gas

P V = m R T

For air R=0.287 KJ/kg.K

P V = m R T

120 x 0.5 = m x 0.287 x 300

m=0.696 kg

`Q=mC_v\Delta T`

Lets take final temperature of air is T

Now by putting the values

`Q=mC_v\Delta T`

`20=0.696* 0.71* (T-300)`

T=340. 47 K

So the final temperature of air will be 340.47 K.