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Serggg [28]
2 years ago
15

Soda pop cane are made of aluminum having thickness of 0.12 mm. They are typically kept at a temperature of 4°C in refrigerator.

Such cane has height of 4.83 inch and diameter of 2.6 inch. Suppose you bought such a cold soda pop and brought into the class where the temperature is 25°C.
Required:
Calculate the rate of heat transfer of the coke cane to the surroundings.
Physics
1 answer:
Advocard [28]2 years ago
5 0

Answer:

Explanation:

For flow of heat through conduction , the formula is

Q = KA( T₂ -T₁ ) / d where K is thermal conductivity of material , A is surface area , T₂ - T₁ is temp diff .

Thermal conductivity of aluminum is 205 W /m.s

Surface area of cane = 2π r² + 2π r h where r is radius and h is height of cane .

A = 2π r ( r + h )

= 2 x 3.14 x 1.3 x .0254 ( 1.3 x .0254 + 4.83 x .0254 )

= .20736 ( .033 + .12268 )

A = .03228 m²

thickness d = .12 x 10⁻³ m

Putting the values ,

Q = 205 x .03228 x ( 25 - 4 ) / .12 x 10⁻³

= 1158 x 10³ J /s

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Answer:

The pressure is p_1 = 4051.4 \ Pa

Explanation:

From the question we are told that

     The gauge pressure at the mouth is  p_1

     The radius of the column is  r_2 =  4 \ mm  =  0.004 \ m

    The speed of the liquid outside the body is  v_2 =  3.1 \ m/s

      The area of the column is  A_2

       The area inside the mouth A_1 = 10 A_2

Generally according to continuity equation

       v_1 A_1 =  v_2 A_2

=>       v_ 1 = v_2 *  \frac{A_2}{A_1}

=>      v_ 1 = 3.1 *  \frac{1}{10}

=>        v_ 1 = 0.31 \ m/s

So

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=>   r_1 = 10 * r_2

substituting values

        r_1 = 10 * 0.004

        r_1 =0.04 \ m

Now the height of inside the mouth is  h_1 =  d =  2r_1 =  2* 0.04 =  0.08\ m

Now the height of the column is  h_2 =  d =  2r_2 =  2* 0.004 =  0.008\ m

Generally according to Bernoulli's  equation

        p_1 =  [\frac{1}{2}  \rho v_2^2 + h_2 \rho g +p_2] -[\frac{1}{2} \rho * v_1^2 + h_1 \rho g ]

Now  \rho =  1000 \ kg m^{-3} which is the density of water

        p_2 is the gauge pressure of the atmosphere which is  zero

 So

       p_1 =  [(0.5 * 1000 * (3.1)^2) +(0.008 * 1000 * 9.8) + 0]-

                                                  [(0.5 * 1000 * 0.31^2) + (0.08*1000 * 9.8)]                          

       p_1 = 4051.4 \ Pa

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A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radiu
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Answer:

1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

Explanation:

According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.

As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :

Q₁ = ∫ ρ dV

Here dV is the volume element of sphere of radius r.

Q₁ = ρ x 4π x ∫ r² dr

The limit of integration is from 0 to r as r is less than R.

Q₁ = (4π x ρ x r³ )/3

But volume charge density, ρ = \frac{3Q}{4\pi R^{3} }

So, Q_{1} = \frac{Qr^{3} }{R^{3} }

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Substitute the value of Q₁ in the above equation. Hence,

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Answer:

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