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19 February, 11:42

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.

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  1. 19 February, 11:48
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    -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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