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likoan [24]
3 years ago
10

List the step that show how the sun provides the energy for surface ocean current.

Chemistry
2 answers:
Tomtit [17]3 years ago
5 0

1.    The sun’s thermal energy is exhibited from its surface traveling to the ocean floor in the day

2.    As the heat energy is absorbed by the water, the currents move and rise as the ocean waters from the equator heads to the north phase of the planet.

3.    And then the waters from the polar areas heads to the south, while ascending and gushing it becomes heated by the sun.


The sun’s gravity also affects this body of water. Since it has a much more stronger gravitation pull than of earth though its far away hence, high tide and low tide tendencies.



Ede4ka [16]3 years ago
4 0
The heat moves in predictable ways, flowing from warmer objects to cooler ones, until both reach the same temperature. When air is in contact with the ocean at a different temperature than the sea  surface, heat transfer by a conductor. The ocean absorbs and stores energy from the sun and when precipitation falls , it release heat energy in the atmosphere(air)
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Answer:

Total heat required to raise the temperature of water from 45.7°C to 103.5°C

= 249,362.4 J

Explanation:

The Heat required to raise the temperature of 100.0 g of water from 45.7°C to 103.5°C will be a sum of;

- The heat required to raise the 100 g of water from 45.7°C to water's boiling point of 100°C

- The Heat required to vaporize the 100 g of water at its boiling point

- The Heat required to raise the temperature of this vapour from 100°C to 103.5°C

1) The heat required to raise the 100 g of water from 45.7°C to water's boiling point of 100°C

Q = mCΔT

m = 100 g

C = 4.18 J/g.°C

ΔT = change in temperature = (100 - 45.7) = 54.3°C

Q = 100 × 4.18 × 54.3 = 22,697.4 J

2) The Heat required to vaporize the 100 g of water at its boiling point

Q = mL

m = 100 g

L = ΔHvaporization = 2260 J/g

Q = mL = 100 × 2260 = 226,000 J

3) The Heat required to raise the temperature of this vapour from 100°C to 103.5°C

Q = mCΔT

m = 100 g

C = 1.90 J/g.°C

ΔT = change in temperature = (103.5 - 100) = 3.5°C

Q = 100 × 1.9 × 3.5 = 665 J

Total heat required to raise the temperature of water from 45.7°C to 103.5°C

= 22,697.4 + 226,000 + 665

= 249,362.4 J

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