. Ropes made from are typically very weak.
1 answer:
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Answer:
9500 kJ; 9000 Btu
Explanation:
Data:
m = 100 lb
T₁ = 25 °C
T₂ = 75 °C
Calculations:
1. Energy in kilojoules
ΔT = 75 °C - 25 °C = 50 °C = 50 K
2. Energy in British thermal units
A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.
Explanation:
From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).
To find the resistance of 260 ft (79.25 m) of size 4 AWG,
R= K * L/ A
K = 0.0214 ohm mm²/m
L = 79.25 m
A = 21.2 mm²
R = 0.0214 *
= 0.0214 * 3.738
= 0.0792 ohm.
Thus the resistance of uncoated copper wire is 0.0792 ohm
Answer:
the highest rate of heat transfer allowed is 0.9306 kW
Explanation:
Given the data in the question;
Volume = 4L = 0.004 m³
V = V
= 0.002 m³
Using Table ( saturated water - pressure table);
at pressure p = 175 kPa;
v = 0.001057 m³/kg
v = 1.0037 m³/kg
u = 486.82 kJ/kg
u 2524.5 kJ/kg
h = 2700.2 kJ/kg
So the initial mass of the water;
m₁ = V/v
+ V
/v
we substitute
m₁ = 0.002/0.001057 + 0.002/1.0037
m₁ = 1.89414 kg
Now, the final mass will be;
m₂ = V/v
m₂ = 0.004 / 1.0037
m₂ = 0.003985 kg
Now, mass leaving the pressure cooker is;
m = m₁ - m₂
m = 1.89414 - 0.003985
m = 1.890155 kg
so, Initial internal energy will be;
U₁ = mu
+ m
u
U₁ = (V/v
)u
+ (V
/v
)u
we substitute
U₁ = (0.002/0.001057)(486.82) + (0.002/1.0037)(2524.5)
U₁ = 921.135288 + 5.030387
U₁ = 926.165675 kJ
Now, using Energy balance;
E - E
= ΔE
QΔt - mh
= m₂u₂ - U₁
QΔt - mh
= m₂u
- U₁
given that time = 75 min = 75 × 60s = 4500 sec
so we substitute
Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675
Q(4500) - 5103.7965 = 10.06013 - 926.165675
Q(4500) = 10.06013 - 926.165675 + 5103.7965
Q(4500) = 4187.690955
Q = 4187.690955 / 4500
Q = 0.9306 kW
Therefore, the highest rate of heat transfer allowed is 0.9306 kW