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

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

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.