Question

A system with 5 kg mass undergoes a process during which there is heat transfer from the system at a rate of 10 kJ per kg, an elevation decrease of 75 m, and an increase in velocity from 10 m/s to 20 m/s. The specific internal energy decreases by 10 kJ/kg. Determine the work transfer during the process.

Answer 1

2.925 kJ

Step-by-step explanation:

Let us assume g = 9.8 m/s²

There is energy transfer by heat from the system, hence Q is negatitive

Q = -(10 kJ per kg)(5 kg) = -50 kJ

Since the specific internal energy decreases, hence it is negative.

U = - 10 kJ/kg(5 kg) = -50 kJ

Since there is decrease in elevation, hence the change in elevation should be negative.

ΔPE = mgΔh = 5kg × 9.8 m/s² × -75 m = -3675 J = -3.675 kJ

There is increase in velocity, hence change in kinetic energy is positive:

ΔKE =
`(1)/(2)m(v_f^2-v_i^2)= (1)/(2)5\ kg(20^2-10^2=750\ J=0.75\ kJ`

Energy in - Energy out = ΔE

Q - W = ΔKE + ΔPE + ΔU

W = Q - (ΔKE + ΔPE + ΔU)

W = -50 kJ - (0.75 kJ + (-3.675 kJ) + (-50 kJ))

W = -50 kJ - (0.75 kJ - 3.675 kJ - 50 kJ)

W = -50 kJ - (-52. 925 kJ) = -50 kJ + 52. 925 kJ

W = 2.925 kJ