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3 years ago

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A uniform electric field is pointing in x direction. The magnitude of the electric field is 10 N/C. The filed makes an angle of

30 deg with a rectangular surface of area 2 x10 m Calculate the electric flux crossing the surface.

1 answer:

Ahat [919]3 years ago

8 0

Answer:

- 100 Nm²/C

Explanation:

E = magnitude of electric field along x-direction = 10 N/C

θ = angle made by the direction of electric field with rectangular surface = 30

φ = angle made by the direction of electric field with normal to the rectangular surface = 90 + 30 = 120

A = area of the rectangular surface = 2 x 10 = 20 m²

Φ = Electric flux crossing the surface

Electric flux is given as

Φ = E A Cosφ

Φ = (10) (20) Cos120

Φ = - 100 Nm²/C

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An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su

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Answer: A. 23.59 minutes.

B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

<em>Latent heat </em><em>of fusion </em>is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, <u>given that</u>:

mass of ice, m= 0.555 kg

temperature of ice, T= -16.6°C

rate of heat transfer, $q=820 J.min^{-1}$

specific heat of ice, $c_{i}= 2100 J.kg^{-1}.K^{-1}$

latent heat of fusion of ice, $L_{i}=334\times10^{3}J.kg^{-1}$

<u>Asked:</u>

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

<u>We have the formula:</u>

$Q=mc\Delta T$

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

$Q= 0.555\times2100\times16.6$

$Q= 19347.3 J$

Now, we require 19347.3 joules of heat to bring the ice to 0°C and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

$t=\frac{Q}{q}$

$=\frac{19347.3}{820}$

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

<em>Now comes the concept of Latent heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.</em>

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

<u>We have the equation:</u> latent heat, $Q_{L}= mL_{i}$

$Q_{L}= 0.555\times334\times10^{3}= 185370 J$

<u>Now the time required for supply of 185370 J:</u>

$t=\frac{Q_{L}}{q}$

$t=\frac{185370}{820}$

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

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3 years ago