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jarptica [38.1K]
2 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]2 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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Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 20 kW of shaft power. The free surface of t
kifflom [539]

Answer:

6.5 kW

Explanation:

Input power = 20 kW = 20000 W

h = 45 m

Volume flow per second = 0.03 m^3 /s

mass flow per second = volume flow per second x density of water

                                    = 0.03 x 1000 = 30 kg/s

Output power = m g h / t = 30 x 10 x 45 = 13500 W

Power converted in form of heat = Input power - Output power

                                                       = 20000 - 13500 = 6500 W = 6.5 kW

Thus, mechanical power converted into heat is 6.5 kW.

7 0
2 years ago
A body A of mass 1.5kg, travelling along the positive x-axis with speed 4.5m/s, collides with another body B of mass 3.2kg which
Pie
Momentum before the collision
x-direction:
p = m₁v₁ = 1.5 * 4.5 = 6.75
x-direction:
p = 0

momentum after the collision is conserved:
x-direction:
p = 6.75 = m₁v₁ + m₂v₂ = 1.5 * 2. 1* cos -30° + 3.2 * v₂*cos θ
y-direction:
p = 0 = m₁v₁ + m₂v₂ = 1.5 * 2.1 * sin -30° + 3.2 * v₂ * sin θ

Solve the two equations for v₂ and θ.

5 0
1 year ago
An object is 6.0 cm in front of a converging lens with a focal length of 10 cm.Use ray tracing to determine the location of the
Masja [62]

As we know that

\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}

here we know that

d_0 = 6 cm

f = 10 cm

now from above equation we have

\frac{1}{d_i} + \frac{1}{6} = \frac{1}{10}

d_i = -15 cm

so image will form on left side of lens at a distance of 15 cm

This image will be magnified and virtual image

Ray diagram is attached below here

8 0
2 years ago
(b) If the straight-line distance from her home to the university is 10.3 km in a direction 25.0° south of east, what was her av
svet-max [94.6K]

Answer:

The time taken is missing in the question. The time is 18 minutes.

The answer is 34.3 km/hr

Explanation:

Average velocity is the speed or the velocity which is required to cover a distance in a time interval.

The time taken is = 18 min

                            = 18/60 hours

The distance from the university to her home is = 10.3 km

Therefore, the average velocity is = displacement/time taken

                                                         = 10.3 / (18/60) km/hr

                                                         = 34.3 km/hr

Hence, the average velocity is 34.3 km/hr

7 0
2 years ago
Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.
kramer

Answer:

115 ⁰C

Explanation:

<u>Step 1:</u> The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

q_{1} +q_{2} =-q_{3} -----eqution 1

where,

q_{1} is the heat absorbed by the solid at 0⁰C

q_{2} is the heat absorbed by the liquid at 0⁰C

q_{3} the heat lost by the warmer water sample

Important equations to be used in solving this problem

q=m *c*\delta {T}, where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

\delta {T} is change in temperature

Again,

q=n*\delta {_f_u_s} -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

<u>Step 2:</u> calculate how many moles of water you have in the 100.0-g sample

=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O

<u>Step 3: </u>calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ

This means that equation (1) becomes

79.13 KJ + q_{2} = -q_{3}

<u>Step 4:</u> calculate the final temperature of the water

79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}

Substitute in the values; we will have,

79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})

79.13 kJ + 990.66J* (T_{f}-218}) = -1463J*(T_{f}-100})

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* (T_{f}-218}) = -1.463KJ*(T_{f}-100})

79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3

collect like terms,

2.45366T_{f} = 283.133

∴T_{f} = = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

6 0
2 years ago
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