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A transfer by work from the system, an elevation decrease of 50 m, and an increase in velocity from 15 m/s to 30 m/s. The specific internal energy, u, decreases by 5 kJ/kg and the acceleration of gravity is constant at 9.7 m/s². Determine the satellite’s heat transfer in the process. Answer: magnitude is about 50 kJ.

Respuesta :

Answer:

-50 kJ.

Explanation:

W = 0.147 * 10

= 1.47 kJ

ΔU = -5 * 10

= - 50 kJ

Potential energy, P = m*g*h

= 50*9.7*(-50)

= -4850 J

= -4.85 kJ

Kinetic energy, K = 1/2 * m * (vf^2 - vi^2)

= 1/2 * 10 * (30^2 - 15^2)

= 3375 J

= 3.375 kJ

ΔE = Ein - Eout

Q - W = ΔK + ΔP + ΔU

Q = 1.47 + 3.375 - 4.85 - 50

= -50.005 kJ

= -50 kJ.

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