Answer:
Fuel efficiency for highway = 114.08 miles/gallon
Fuel efficiency for city = 98.79 miles/gallon
Explanation:
1 gallon = 3.7854 litres
1 mile = 1.6093 km
Let's first convert the efficiency to km/gallon:
48.5 km/litre = (48.5 * 3.7854) km/gallon
48.5 km/litre = 183.5919 km/gallon (highway)
42.0 km/litre = (42.0 * 3.7854) km/gallon
42.0 km/litre = 158.9868 km/gallon (city)
Next, we convert these to miles/gallon:
183.5919 km/gallon = (183.5919 / 1.6093) miles/gallon
183.5919 km/gallon = 114.08 miles/gallon (highway)
158.9868 km/gallon = (158.9868 /1.6093) miles/gallon
158.9868 km/gallon = 98.79 miles/gallon (city)
Answer:
0.948 Btu
Explanation:
1 Btu = 1055 J so 
= 0.948 Btu
Answer:
The final temperature in the vessel after the resistor has been operating for 30 min is 111.67°C
Explanation:
given information:
mass, m = 3 kg
initial temperature, T₁ = 40°C
current, I = 10 A
voltage, V = 50 V
time, t = 30 min = 1800 s
Heat for the system because of the resistance is
Q = V I t
where
V = voltage (V)
I = current (A)
t = time (s)
Q = heat transfer to the system (J)
so,
Q = V x I x t
= 50 x 10 x 1800
= 900000
= 9 x 10⁵ J
the heat transfer in the closed system is
Q = ΔU + W
where
U = internal energy
W = work done by the system
thus,
Q = ΔU + W
9 x 10⁵ = ΔU + 0, W = 0 because the tank is a well-insulated and rigid.
ΔU = 9 x 10⁵ J = 900 kJ
then, the energy change in the system is
ΔU = m c ΔT
ΔT = ΔU / m c, c = 4.186 J/g°C
= 900 / (3 x 4.186)
= 71.67°C
so,the final temperature (T₂)
ΔT = T₂ - T₁
T₂ = ΔT + T₁
= 71.67°C + 40°C
= 111.67°C
Answer:
The pumping power per ft of pipe length required to maintain this flow at the specified rate 0.370 Watts
Explanation:
See calculation attached.
- First obtain the properties of water at 60⁰F. Density of water, dynamic viscosity, roughness value of copper tubing.
- Calculate the cross-sectional flow area.
- Calculate the average velocity of water in the copper tubes.
- Calculate the frictional factor for the copper tubing for turbulent flow using Colebrook equation.
- Calculate the pressure drop in the copper tubes.
- Then finally calculate the power required for pumping.
