Apiston–cylinder device contains 5 kg of air at 400 kpa and 30oc. during a quasi-equilibrium isothermal expansion process, 15 kj of boundary work is done by the system, and 3 kj of paddle-wheel work is done on the system. determine the amount and direction of the heat transfer during this process.

Q= 12 KJ

Heat enters to the system.

Explanation:

Given that

m= 5 kg

P₁ = 400 KPa

T₁ = 30⁰C = 303 K

Work done by the system ,W₁= 15 KJ

Work is done on the system W₂ = - 3 KJ'

Given that expansion is isothermal ,that is why temperature of the gas will remain constant.We know that internal energy of the ideal gas only depends on the temperature of the gas.That is why change in the internal energy of the gas will be zero because change in the temperature is zero.

From first law of thermodynamics

Q = W + ΔU

Q= =Heat ,W= Net Work  ,ΔU =Change in the internal energy

Here ΔU = 0

Q= W

Q= 15 - 3 KJ

Q= 12 KJ

Heat enters to the system.

Note -

Work done on the system will be taken as negative and work done by the system taken as positive.

Heat added to the system will be taken as positive heat leaving from the system taken as negative.

answer

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explanation:

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever