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## 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 eleva

Question

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

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Physics
3 years
2021-08-15T20:32:57+00:00
2021-08-15T20:32:57+00:00 2 Answers
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## Answers ( )

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.

Answer:

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