Answered

A well insulated and perfectly sealed room is installed with 4 HP fan to circulate air for 1.5 hours. Determine the increase in internal energy of the air answer is 1.61 x10^7 J

Verified Answer :

"Increase in internal energy = energy consumed by fan " as the room is perfectly insulated and sealed.

energy consumed by fan = power x time
power of fan = 4HP = 4 x 745.7 W = 2982.8 W
time = 1.5 hours = 1.5 x 60 x 60 = 5400 s

 energy consumed by fan = 2982.8 x 5400 = 16107120 J = 1.61x10^7 J

Increase in internal energy is 1.61x10^7 J


Cricetus

The increase in internal energy will be "1.61×10⁷ J".

Given:

Power,

  • \(P = 4 \ HP\)

or,

           \(= 4\times 745.7\)

           \(= 2982.8 \ W\)

Time,

  • \(T = 1.5 \ hours\)

or,

           \(= 1.5\times 60\times 60\)

           \(= 5400 \ s\)    

Now,

The increase in the internal energy:

= \(Power\times Time\)

= \(2982.8\times 5400\)

= \(1.61\times 10^7 \ J\)

Thus the response above is right.

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