Respuesta :
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
