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jok3333 [9.3K]
3 years ago
8

State how much energy is transferred in each of the following cases: 2 grams of steam at 100 degrees Celsius condenses to water

at 100 degrees Celsius.
2 grams of boiling water at 100 degrees Celsius cools to ice water at 0 degrees Celsius.
2 grams of ice water at 0 degrees Celsius freezes to ice at 0 degrees Celsius.
2 grams of steam at 100 degrees Celsius turns to ice at 0 degrees Celsius.
Physics
1 answer:
dusya [7]3 years ago
3 0

Answer:

Explanation:

When 2 gms of steam condenses to water at 100 degree latent heat of vaporization is releases which is calculated as follows

Heat released = mass x latent heat of vaporization

= 2 x 2260 = 4520 J

When 2 gms of water  at 100 degree is cooled to ice water at zero degree  heat  is releases which is calculated as follows

Heat released = mass x specific heat x( 100-0)

= 2 x 4.2 x 100 = 840 J

When 2 gms of water at zero degree  condenses to ice at zero degree latent heat of fusion  is releases which is calculated as follows

Heat released = mass x latent heat of fusion

= 2 x 334 = 668 J

When 2 grams of steam at 100 degrees Celsius turns to ice at 0 degrees Celsius heat released will be sum of all the heat released as mentioned above ie

4520 + 840 +668 = 6028 J

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Answer:

Answer:

A. - 0.017N. It acts to the left.

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Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

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Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

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F = -0.017N

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B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

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The correct answer is C) Outward Pressure and Inward gravity                                                        

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