Question

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.

Answer 1

The change in kinetic energy is 4.3875 kJ

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

Step-by-step 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