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jarptica [38.1K]
3 years ago
15

If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2

.5 cm thick and the thermal conductivity may be taken as 0.2 W/m K, what is the temperature difference across the material?
Physics
1 answer:
kaheart [24]3 years ago
5 0

Answer:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

Explanation:

For this case we can use the Fourier Law of heat conduction given by the following equation:

Q = -kA \frac{\Delta T}{\Delta x}   (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 represent the cross sectional area

Q= 3KW represent the rate of heat transfer

\Delta T is the temperature of difference that we want to find

\Delta x=2.5 cm =0.025 m represent the thickness of the material

If we solve \Delta T in absolute value from the equation (1) we got:

\Delta T =\frac{Q \Delta x}{Ak}

First we convert 3KW to W and we got:

Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W

And we have everything to replace and we got:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

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3 years ago
A total of 25.6 kJ of heat energy is added to a 5.46 L sample of helium at 0.991 atm. The gas is allowed to expand against a fix
bagirrra123 [75]

Answer:

(a) W = 1329.5 J = 1.33 KJ

(b) ΔU = 24.27 KJ

Explanation:

(a)

Work done by the gas can be found by the following formula:

W = P\Delta V

where,

W = Work = ?

P = constant pressure = (0.991 atm)(\frac{101325\ Pa}{1\ atm}) = 100413 Pa

ΔV = Change in Volume = 18.7 L - 5.46 L = (13.24 L)(\frac{0.001\ m^3}{1\ L}) = 0.01324 m³

Therefore,

W = (100413 Pa)(0.01324 m³)

<u>W = 1329.5 J = 1.33 KJ</u>

<u></u>

(b)

Using the first law of thermodynamics:

ΔU = ΔQ - W (negative W for the work done by the system)

where,

ΔU = change in internal energy of the gas = ?

ΔQ = heat added to the system = 25.6 KJ

Therefore,

ΔU = 25.6 KJ - 1.33 KJ

<u>ΔU = 24.27 KJ</u>

3 0
3 years ago
point) A circular swimming pool has a diameter of 12 m. The circular side of the pool is 3 m high, and the depth of the water is
Sergio [31]

Answer:

(a) 86.65 J

(b) 149.65 J

Solution:

As per the question:

Diameter of the pool, d = 12 m

⇒ Radius of the pool, r = 6 m

Height of the pool, H = 3 m

Depth of the pool, D = 2.5 m

Density of water, \rho_{w} = 1000\ kg//m^{3}

Acceleration due to gravity, g = 9.8\ m/s^{2}

Now,

(a) Work done in pumping all the water:

Average height of the pool, h = \frac{H + D}{2}

h = \frac{3 + 2.5}{2} = 2.75\ m

Volume of water in the pool, V = \pi r^{2}h = \pi \times 6^{2}\times 2.75 = 311.02\ m^{3}

Mass of water, m_{w} = \frac{\rho_{w}}{V}

m_{w} = \frac{1000}{311.02} = 3.215\ kg

Work done is given by the potential energy of the water as:

W = m_{w}gh = 3.215\times 9.8\times 2.75 = 86.65\ J

(b) Work done to pump all the water through an outlet of 2 m:

Now,

Height, h = 2.75 + 2 = 4.75

Work done,W = m_{w}gh = 3.215\times 9.8\times 4.75 = 149.65\ J

7 0
3 years ago
The acceleration due to gravity on the surface of Venus is 8.83 m/s2. An object with a mass of 5.23 kg has what weight on Venus?
Marat540 [252]
Weight = m times g = 5.23 times 8.83 = 46.18 N
6 0
3 years ago
a baseball is hit 3 feet above ground level at 100 feet per second and at an angle of 45 with respect to the ground. (g=32 feet/
LiRa [457]

Answer:

hmax=81ft

Explanation:

Maximum height of the object is the highest vertical position along its trajectory.

The vertical velocity is equal to 0 (Vy = 0)

0=V_{y}-g*t=v_{0}*sin(\alpha)-g*th\\

we isolate th (needed to reach the maximum height hmax)

th = \frac{v_{0}*sin(\alpha)}{g}

The formula describing vertical distance is:

y = Vy * t-g* t^{2} / 2

So, given y = hmax and t = th, we can join those two equations together:

hmax = Vy* th-g*th^{2}/2

hmax =Vo^{2}*sin(\alpha )^{2}/(2*g)

if we launch a projectile from some initial height h all you need to do is add this initial elevation

hmax =h+Vo^{2}*sin(\alpha)^{2}/(2*g)

hmax =3+100^{2}*sin(45)^{2}/(2 * 32)=81 ft

6 0
3 years ago
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