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vesna_86 [32]
3 years ago
12

Hillsdale and Cordera are the same distance from the equator, and both are near the ocean. Is Cordera warmer, colder, or the sam

e temperature as Hillsdale? I saw someone else ask this but I am still looking for the answer. I need it A.S.A.P since it is due today. Thx :)
Chemistry
2 answers:
SIZIF [17.4K]3 years ago
5 0

Answer:

Cordera can be warmer or colder than Hillsdale but can never be of the same temperature.

Explanation:

The reasons depend on the direction of the air current. And the position of the two cities to the equator. Because energy from the sun is transferred to Earth’s surface. Some of that energy is then transferred to the air above the surface. The closer a location is to the equator, the more energy it receives from the sun. Therefore, a location’s air temperature is affected by its distance from the equator. This is not the only cause for cooling.

Water moving from the equator is warmer than would be expected based on latitude, and so is warmer than the air it passes. Water moving from a pole is colder than would be

expected based on latitude and so is colder than the air it passes. Energy transfers from warmer substances to colder substances. Warmer air transfers energy to cooler currents and warmer currents transfer energy to cooler

air.

hram777 [196]3 years ago
3 0

Answer:

They'll both have the same temperature.

Explanation:

Places at the same latitude may have very different climates if one is on a coast and one is inland. On the coast, the climate is influenced by warm moist air from the ocean, but farther inland, the climate is influenced by cold or hot air from the land. Since they're both beside the ocean, I'd say they'll have the same temperature.

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hammer [34]

Answer:

(Q1) 9.42 kJ.

(Q2) 1.999 kJ

Explanation:

Heat: This is a form of Energy that brings about the sensation of warmth.

The S.I unit of Heat is Joules (J).

The heat of a body depend on the mass of the body, specific heat capacity, and temperature difference. as shown below

Q = cm(t₂-t₁) ........................ Equation 1

(Q1)

Q = cm(t₂-t₁)

Where Q = amount of heat absorbed, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature.

Given: m = 229.1 g = 0.2991 kg, t₁ = 25.0 °C, 32.50 °C

Constant: c = 4200 J/kg.°C

Substituting into equation 1

Q = 0.2991×4200(32.5-25)

Q = 1256.22(7.5)

Q = 9421.65 J

Q = 9.42 kJ.

Hence the heat absorbed = 9.42 kJ

(Q2)

Q = cm(t₂-t₁)

Where Q = amount of heat required, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature.

Given: m = 34 g = 0.034 kg, t₁ = 9 °C, t₂ = 23 °C

Constant: c = 4200 J/kg.°C

Q = 0.034×4200(23-9)

Q = 142.8(14)

Q = 1999.2 J

Q = 1.999 kJ.

Thus the Heat required = 1.999 kJ

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