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brilliants [131]
3 years ago
10

What is the thermal energy contained within a cubic kilometre of granite (in barrels of oil equivalent) for every one degree cha

nge in temperature? Use the following parameters in your calculation: Specific heat of granite = 790 (J/kgC) Density of granite = 2700 kg/m3 Energy in 1 barrel of oil = 6.118e9 J
Engineering
1 answer:
AnnZ [28]3 years ago
4 0

Answer:

348643.34 barrels of oil

Explanation:

Given;

Volume of the granite = 1 km³ = 1000³ m³

Specific heat of granite, C = 790 J/kg.°C

Density of granite = 2700 kg/m³

Energy in 1 barrel of oil = 6.118 × 10⁹ J

For every 1° change in temperature, ΔT = 1°C

Now,

The thermal energy stored is given as;

Thermal energy = mCΔT

where, m is the mass

the mass of 1 km³ of granite = Density × Volume

or

the mass of 1 km³ of granite = 2700 × 1000³ = 27 × 10¹¹ Kg

therefore,

Thermal energy = 27 × 10¹¹ × 790 × 1

or

Thermal energy = 21330 × 10¹¹ J

hence,

the thermal energy in terms of barrels of oil

= Total thermal energy / Energy stored in 1 barrel of oil

= 21330 × 10¹¹ J / ( 6.118 × 10⁹ J per barrel )

= 348643.34 barrels of oil

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2 years ago
If there are 16 signal combinations (states) and a baud rate (number of signals/second) of 8000/second, how many bps could I sen
Mice21 [21]

Answer:

32000 bits/seconds

Explanation:

Given that :

there are 16  signal combinations (states) = 2⁴

bits  n = 4

and a baud rate (number of signals/second) = 8000/second

Therefore; the number of bits per seconds can be calculated as follows:

Number of bits per seconds = bits  n × number of signal per seconds

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2 years ago
If you were driving the blue Prius in the situation pictured above, explain why the red Mustang should be given right-of-way at
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Answer:

The blue Prius, because the Mustang arrived almost in the same time. And when you arrive in an intersection at the same time of other vehicle you need yield for the car on your right if the car is on your left you have the right of way.

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4 0
2 years ago
A welding rod with κ = 30 (Btu/hr)/(ft ⋅ °F) is 20 cm long and has a diameter of 4 mm. The two ends of the rod are held at 500 °
SOVA2 [1]

Answer:

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

Explanation:

Thermal Conductivity=k=30  (Btu/hr)/(ft ⋅ °F)= \frac{30}{3600} (Btu/s)/(ft.F)=8.33*10^{-3}  (Btu/s)/(ft.F)

Thermal Conductivity is SI units:

k=30(Btu/hr)/(ft.F) * \frac{1055.06}{3600*0.3048*0.556} \\k=51.88 W/m.K

Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft

Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft

T_1=500 C=932 F

T_2=50 C= 122 F

Part A:

In Joules (J)

A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J

In Btu:

A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu

PArt B:

At midpoint Length=L/2=0.1 m

Q=\frac{k*A*(T_1-T_2)}{L}

On rearranging:

T_2=T_1-\frac{Q*L}{KA}

T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C

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2 years ago
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Paul [167]

Answer:b

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6 0
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