A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per kg, an elevation decrease of 50 m, and an increase in velocity 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 9.7 m/s 2. Determine the heat transfer for the process, in kJ

Given Information:

Mass = m = 10 kg

Work done by system = W = 0.147 kJ/kg

Elevation = Δh = - 50m (minus sign because decreases)

Final velocity = v₂ = 30 m/s

Initial velocity = v₁ = 15 m/s

Internal energy = ΔU = - 5 kJ/kg (minus sign because decreases)

Required Information:

Heat transfer = Q = ?

Answer:

Heat transfer ≈ - 50 kJ

Explanation:

We know that heat transferred to the system is equal to

Q = W + ΔKE + ΔPE + ΔU

The Work done by the system in Joules is

W = 0.147 kJ/kg * 10 kg

W = 1.47 kJ

The change in kinetic energy is given by

ΔKE = ½m (v₂ - v₁) ²

ΔKE = ½*10 * (30 - 15) ²

ΔKE = 5 * (675)

ΔKE = 3.375 kJ

The change in potential energy is given by

ΔPE = mgΔh

Where g is 9.7 m/s²

ΔPE = 10*9.7*-50

ΔPE = - 4.85 kJ

The Internal energy in Joules is

ΔU = - 5 kJ/kg * 10 kg

ΔU = - 50 kJ

Therefore, the heat transfer for the process is

Q = W + ΔKE + ΔPE + ΔU

Q = 1.47 + 3.375 - 4.85 - 50

Q = 50.005 kJ

Q ≈ - 50 kJ

Bonus:

The negative sign indicates that the heat was transferred from the system. It would have been positive if the heat was transferred into the system.

Heat transfer for the process is - 52.255 kJ.

Explanation:

Heat transfer for the process is a summation of the work done by the system, change in kinetic energy, change in potential energy and change in internal energy. That is, Q = W + ∆K. E + ∆P. E + ∆U

Mass (m) of the system = 10 kg

W = 0.147 kJ/kg = 0.147 kJ/kg * 10 kg = 1.47 kJ

∆K. E = 1/2m (v2 - v1) ^2

V2 = 30 m/s

V1 = 15 m/s

∆K. E = 1/2*10 (30 - 15) ^2 = 1,125 J = 1,125/1000 = 1.125 kJ

∆P. E = mg (h2 - h1)

g = 9.7 m/s^2

(h2 - h1) is change in elevation. It decreased by 50 m. Therefore, (h2 - h1) = - 50 m

∆P. E = 10*9.7*-50 = - 4850 J = - 4850/1000 = - 4.85 kJ

∆U (change in velocity) decreases by 5 kJ/kg. Therefore, ∆U = - 5 kJ/kg = - 5 kJ/kg * 10 kg = - 50 kJ

Q = 1.47 kJ + 1.125 kJ + (-4.85 kJ) + (-50 kJ) = 1.47 kJ + 1.125 kJ - 4.85 kJ - 50 kJ = - 52.255 kJ