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Paraphin [41]
4 years ago
8

A 360-g metal container, insulated on the outside, holds 180.0 g of water in thermal equilibrium at 22.0°C. A 24.0-g ice cube, a

t the melting point, is dropped into the water, and when thermal equilibrium is reached the temperature is 15.0°C. Assume there is no heat exchange with the surroundings. For water, the specific heat capacity is 4190 J/kg ∙ K and the heat of fusion is 3.34 × 105 J/kg. What is the specific heat capacity of the metal of the container?
Physics
1 answer:
matrenka [14]4 years ago
3 0

Answer: specific heat capacity of the metal of the container

Cm=1143.1J/kg ∙ K

Explanation:

Step one

Given the following data

Mass of metal container Mm=360g

Mass of water Mw=180g

Mass of ice Mice=24g

T1=22°C

T2=15°C

specific heat capacity of water

Cw = 4190 J/kg ∙ K

the heat of fusion of ice

Lf= 3.34 × 10^5 J/kg.

specific heat capacity of the metal of the container Cm=?

Step two

Heat loss by ice cube =

heat gained by water + heat gained by metal container

I.e

Mice*Lice= MwCw(T1-T2)w

+MmCm(T1-T2)

Substituting our data we have

24*3.4*10^5=180*4190*(22-15)

+360*Cm*(22-15)

8160000=754200*7+2520Cm

8160000=5279400+2520Cm

8160000-5279400 =2520Cm

2880600=2520Cm

Divide both side by 2520

2880600/2520=Cm

Cm=1143.1J/kg ∙ K

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The density of mobile electrons in copper metal is 8.4 × 1028 m-3. Suppose that i= 4.4 × 1018 electrons/s are drifting through a
neonofarm [45]

Answer: 405.3 minutes

Explanation: In order to explain this problem we have to use the following:

Fisrtly we calculate the volume of the wire, this is given by:

Vwire=π*r^2*L where r and L are the radius and L the length of teh wire, respectively.

Vwire=π*1.25*10^-3*0.26=1.27*10^-6 m^3

then the number of the total electrons in tthe wire volume is given by;

n° electrons in the wire=ρ*Vwire=8.4*10^28*1.27*10^-6 m^3=1.07 *10^23

Finally, considering the current in the wire equal to 4.4*10^18 electrons/s

the time consuming to extract all the electrons from the wire is given by:

t= total electrons in the wire/ current=1.067*10^23/4.4*10^18=24,318 s

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4 0
3 years ago
A large solar panel on a spacecraft in Earth orbit produces 1.0 kW of power when the panel is turned toward the sun. What power
Mandarinka [93]

Answer:

e*P_s = 11 W

Explanation:

Given:

- e*P = 1.0 KW

- r_s = 9.5*r_e

- e is the efficiency of the panels

Find:

What power would the solar cell produce if the spacecraft were in orbit around Saturn

Solution:

- We use the relation between the intensity I and distance of light:

                                  I_1 / I_2 = ( r_2 / r_1 ) ^2

- The intensity of sun light at Saturn's orbit can be expressed as:

                                  I_s = I_e * ( r_e / r_s ) ^2

                                  I_s = ( 1.0 KW / e*a) * ( 1 / 9.5 )^2

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- We know that P = I*a, hence we have:

                                  P_s = I_s*a

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Hence,                       e*P_s = 11 W

3 0
3 years ago
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A light wave that hits the surface of a pool gets refracted and gives us an apparent image of the surface of the pool, following the concepts of refraction.

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Let’s recall the concept of refraction when a light wave passes from medium of rarer to denser. There is a change in the speed of light while travelling from medium of rarer to denser.

There can be a change in the direction as well. This property is known as “Refraction” and the best example to see refraction is watching the surface of a clean pond, lake or pool.

When the light travels from a rarer medium (air) to a denser medium (water), it changes its angle of direction and gets refracted and hit to our eye lenses. With this, we see the surface of the pool at a changed angle and it seems to be a bit shallow than its original depth.

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Explanation:

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