Answer:
c. less than 60 mi/h
Explanation:
To calculate the average speed of the bus, we need to calculate the total distance traveled by the bus, as well as the total time of travel of the bus.
Total Distance Traveled = S = 100 mi + 100 mi
S = 200 mi
Now, for total time, we calculate the times for both speeds from A to b and then B to C, separately and add them.
Total Time = t = Time from A to B + Time from B to C
t = (100 mi)/(50 mi/h) + (100 mi)(70 mi/h)
t = 2 h + 1.43 h
t = 3.43 h
Now, the average speed of bus will be given as:
Average Speed = V = S/t
V = 200 mi/3.43 h
<u>V = 58.33 mi/h</u>
It is clear from this answer that the correct option is:
<u>c. less than 60 mi/h</u>
Answer:
2.5 is the required details
Answer:
(a) The Final Temperature is 315.25 K.
(b) The amount of mass that has entered 0.5742 Kg.
(c) The work done is 56.52 kJ.
(d) The entrophy generation is 0.0398 kJ/kgK.
Explanation:
Explanation is in the following attachments.
Answer:
(a)
<em>d</em>Q = m<em>d</em>q
<em>d</em>q = 
<em>d</em>T
= (T₂ - T₁)
From the above equations, the underlying assumption is that remains constant with change in temperature.
(b)
Given;
V = 2L
T₁ = 300 K
Q₁ = 16.73 KJ , Q₂ = 6.14 KJ
ΔT = 3.10 K , ΔT₂ = 3.10 K for calorimeter
Let 
be heat constant of calorimeter
Q₂ = ΔT
Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂
Q₁ - Q₂ = m ΔT
number of moles of n-C₆H₁₄, n = m/M
ρ = 650 kg/m³ at 300 K
M = 86.178 g/mol
m = ρv = 650 (2x10⁻³) = 1.3 kg
n = m/M => 1.3 / 0.086178 = 15.085 moles
Q₁ - Q₂ = m ' ΔT
= (16.73 - 6.14) / (15.085 x 3.10)
= 0.22646 KJ mol⁻¹ k⁻¹
