Aclosed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kj per kg, an elevation decrease of 50 m, and an increase in velocity from 15 m/s to 30 m/s. the specific internal energy decreases by 5 kj/kg and the acceleration of gravity is constant at 9.7 m/s^2. determine the heat transfer for the process, in kj.

50.005 kJ of heat is transferred by the system

Solution:

As per the question:

Mass of the closed system, m = 10 kg

Decrease in elevation, d = 50 m

Initial velocity, v = 15 m/s

Final velocity, v' = 30 m/s

Change in internal energy, dU = - 5 kJ/kg = - 5000(10 kg) = - 50000 J

As per the first thermodynamics' law:

dQ = dW + (dPE +dKE + dU)            (1)

where

dQ = heat transfer change

dPE +dKE + dU = dE = change in the energy of the system

PE = Potential Energy

KE = Kinetic Energy

U = Internal energy

Now,

dPE = mgd = -

(Since, the elevation decreases and hence PE decreases)

dKE =

Work done, dW =

Therefore, using the respective energy values in eqn (1):

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation: