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1 year ago

100 PTS+BRAINILIEST: Why does it get colder as you go higher through the atmosphere, if you are getting closer to the sun??

Chemistry

2 answers:

sattari [20]1 year ago

8 0

Answer/Explanation:

Heat from the sun hits the ground and is absorbed. The higher you go, it gets more colder due to the fact that the air doesn't hold onto the radiation as it goes straight through the ground, which is why the top of mountains are very cold, and people are able to die in minutes.

Nutka1998 [239]1 year ago

7 0

As you go higher, there are less air molecules and matter therefore meaning there is a lower pressure pushing down on you. Plus, when it is lower that means the particles will do some work meaning it will lose energy and when energy decreases so does the temperature and we define temperature as the average energy of the particles, in simpler words: When the pressure gets lower, the temperature also will get lower, which is why the higher you are the air temperature will get lower, and the lower you are the air temperature will get higher.

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3 years ago

It takes 49.0J to raise the temperature of an 11.5g piece of unknown metal from 13.0?C to 24.3?C. What is the specific heat for

svet-max [94.6K]

Answer:

c = 0.377 J/g.°C

c = 0.2350 J/g.°C

J = 27.3 J

Explanation:

We can calculate the heat (Q) absorbed or released by a substance using the following expression.

Q = c × m × ΔT

where,

c: specific heat

m: mass

ΔT: change in the temperature

It takes 49.0J to raise the temperature of an 11.5g piece of unknown metal from 13.0°C to 24.3°C. What is the specific heat for the metal? Express your answer numerically, in J/g.°C

Q = c × m × ΔT

49.0 J = c × 11.5 g × (24.3°C - 13.0°C)

c = 0.377 J/g.°C

The molar heat capacity of silver is 25.35 J/mol.°C. How much energy would it take to raise the temperature of 11.5g of silver by 10.1°C? Express your answer numerically, in Joules. What is the specific heat of silver?

The molar mass of silver is 107.87 g/mol. The specific heat of silver is:

$\frac{25.35J}{mol.\° C} .\frac{mol}{107.87g} =0.2350J/g.\° C$

Q = c × m × ΔT

Q = (0.2350 J/g.°C) × 11.5 g × 10.1°C = 27.3 J

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