A system of mass 13 kg undergoes a process during which there is no work, the elevation decreases by 50 m, and the velocity increases 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 9.8 m/s2. Determine the change in kinetic energy, in kJ, and the amount of energy transfer by heat for the process, in kJ.

The change in kinetic energy is 4.3875 kJ

The amount of energy transferred by heat for the process is -66.98 kJ

Explanation:

Given;

mass of the system, m = 13 kg

change in height, Δh = -50 m

initial velocity, u = 15 m/s

final velocity, v = 30 m/s

change in internal energy per mass, ΔU = -5 kJ/kg

The change in kinetic energy is given by;

ΔK.E = K.E₂ - K.E₁

ΔK.E = ¹/₂mv² - ¹/₂mu²

ΔK.E = ¹/₂m(v² - u²)

ΔK.E = ¹/₂ ₓ 13 (30² - 15²)

ΔK.E = 4387.5 J

ΔK.E = 4.3875 kJ

The amount of energy transferred by heat for the process;

Q = W + ΔP.E + ΔK.E + ΔU

Where;

ΔP.E = mgΔh

ΔP.E = 13 x 9.8 x (-50)

ΔP.E = -6370 J = -6.37 kJ

W = 0

ΔU = -5kJ/kg x 13kg

ΔU = -65 kJ

Q = W + ΔP.E + ΔK.E + ΔU

Q = 0 + (-6.37) + (4.3875) + (-65)

Q = -66.98 kJ

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