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photoshop1234 [79]
3 years ago
14

How much energy in joule is added to a 12 g of sample of aluminum (c=0.897 J/g ◦C) to raise the temperature from 20 ◦C to 45 ◦C?

Engineering
1 answer:
MArishka [77]3 years ago
8 0

Answer:

269.1J

Explanation:

m = 12g

c = 0.897J/g°C

∆T = 45 - 20 = 25°C

H = mc∆T = 12 × 0.897 × 27 = 269.1J

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The air conditioner in a house or a car has a cooler that brings atmospheric air from 30C to 10C, with both states at 101KPa. If
torisob [31]

The rate of heat transfer by the air conditioner using constant specific heat of 1.004kj/kg.K is 15.06 kW.

<h3>What is the rate of heat transfer?</h3>

Rate of heat transfer is the power rating of the machine.

Work done and changes in potential and kinetic energy are neglected since it is a steady state process.

The specific heat in terms of specific heat capacity and temperature change is given as:

q_{out} = Cp(Ti - Te)

q_{out} = 1.004(30 - 10) = 20.08 kJ/kg \\

The rate of heat transfer, is then determined as follows:

  • Qout = flow rate × specific heat

Qout = 0.75 × 20.08 = 15.06 kW

Therefore, the rate of heat transfer by the air conditioner is 15.06 kW.

Learn more about rate of heat transfer at: brainly.com/question/17152804

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6 0
2 years ago
A civil engineer is asked to design a curved section of roadway that meets the following conditions: With ice on the road, when
lianna [129]

Answer:

1. 3.4^{o}

2. 163.3 m

Explanation:

Static friction between road and rubber, μs =0.06

The maximum speed of the car, v = 50 km/h

                                              = (50)(1000/3600) m/s

                                               = 13.89 m/s

The acceleration due to gravity, g = 9.81 m/s^{2}

The frictional force, f = μsN     ...... (1)

The component mg cosθ which balance the normal reaction N

The component mg sinθ acts in an opposite direction to the frictional force f.

        ΣF = mg sinθ-f = 0      ...... (2)

Substitute the equation (1) in equation (2), we get

 ΣF = mgsinθ-μsN = 0

 mgsinθ-μsmgcosθ =0

 μs = sinθ/cosθ

   tanθ = μs

    θ = tan-1( μs) = tan-1(0.06) = 3.4^{o}

(b)The vertical component of the force is

N cosθ = fsinθ+mg

 N cosθ = μsNsinθ+mg

N[cosθ- μs sinθ] = mg     ...... (3)

The horizontal component of the force along the motion of the car is

Nsinθ+fcosθ = ma  (Centripetal acceleration, a = \frac {v^{2}}{r}

  Nsinθ+fcosθ = m(\frac {v^{2}}{r})

   Nsinθ+μsNcosθ = m(\frac {v^{2}}{r})

N[sinθ+μs cosθ] = m(\frac {v^{2}}{r})     ...... (4)    

Dividing the equation (4) with equation (3),

 [sinθ+μscosθ]/[cosθ- μs sinθ] = \frac {v^{2}}{rg}

 cosθ[sinθ/cosθ+μs]/cosθ[1- μs sinθ/cosθ] =\frac {v^{2}}{rg}

[tanθ+μs]/[1-μs tanθ] = \frac {v^{2}}{rg}      

 From part (1), tanθ = μs

 Then the above equation becomes

 \frac {(\mu_s+\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}

\frac {(2\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}

Therefore, the minimum radius of the curvature of the curve is

               r = \frac {v^{2}}{{2 \mu_s/[1-\mu_s^{2}]}g} 

                   = \frac {v^{2}[1-\mu_s^{2}]}{2\mu_s g}

                   = \frac {(13.89 m/s)^{2}[1-(0.06)^{2}]}{(2)(0.06)(9.81)}

                 = 163.3 m

5 0
3 years ago
The formation of faults in Earth's crust is an effect. What causes faults to form in the crust? Global Positioning System sensor
allsm [11]

Answer: The movement of tectonic plates

Explanation:

Tectonic plates are the part of the earth's crust that both the ocean and land rest on. These plates are constantly moving as a result of currents in the mantle.

These movements cause stress on the surface which has the effect of fracturing rocks and thereby creating/ forming faults in the earth's crust. Sometimes faults form when these plates move away from each other and sometimes they are formed when they push into each other.

7 0
2 years ago
Which option distinguishes the members of a software deployment process team most likely involved in the following scenario?
Alchen [17]

Answer:

A local bank, with several branches in three cities, requests changes to its mortgage calculation software.

5 0
2 years ago
Consider a plane composite wall that is composed of two materials of thermal conductivities kA = 0.1 W/m*K and kB = 0.04 W/m*K a
nadya68 [22]

Answer:

q=39.15 W/m²

Explanation:

We know that

Thermal resistance due to conductivity given as

R=L/KA

Thermal resistance due to heat transfer coefficient given as

R=1/hA

Total thermal resistance

R_{th}=\dfrac{L_A}{AK_A}+\dfrac{L_B}{AK_B}+\dfrac{1}{Ah_1}+\dfrac{1}{Ah_2}+\dfrac{1}{Ah_3}

Now by putting the values

R_{th}=\dfrac{0.01}{0.1A}+\dfrac{0.02}{0.04A}+\dfrac{1}{10A}+\dfrac{1}{20A}+\dfrac{1}{0.3A}

R_{th}=4.083/A\ K/W

We know that

Q=ΔT/R

Q=\dfrac{\Delta T}{R_{th}}

Q=A\times \dfrac{200-40}{4.086}

So heat transfer per unit volume is 39.15 W/m²

q=39.15 W/m²

4 0
3 years ago
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