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 = ?

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.