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3 years ago

11

What causes hot air to rise, and cool air to sink?

1 answer:

nataly862011 [7]3 years ago

8 0

Answer:

hot air rises because gases expand as they heat up.When air heats up and expands,its density also increases.The warmer,less dense air effectively floats on top of the colder, dense air below it. This creates a buoyant force that causes warmer air to rise.

cold air sinks because it is heavier as its more dense ( because of closely packed molecules) soits harder for them to move and they absorb less energy. Also, gravity pulls on it more strongly.

Hope it helps :)

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You placed 43.1 g of an unknown metal at 100 °C into a coffee cup calorimeter that contained 50.0 g of water that was initially

nika2105 [10]

Answer :

(a) The heat released by the metal is -312.48 J

(b) The specific heat of the metal is $0.0944J/g^oC$

Explanation :

<u>For part A :</u>

Heat released by the metal = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the metal

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $51.5J/^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_2$ = mass of water = 50.0 g

$\Delta T$ = change in temperature = $(T_{final}-T_{initial})=23.2-22.0=1.2^oC$

Now put all the given values in the above formula, we get:

$q=[(51.5J/^oC\times 1.2^oC)+(50.0g\times 4.184J/g^oC\times 1.2^oC)]$

$q=312.48J$

Thus, the heat released by the metal is -312.48 J

<u>For part B :</u>

$q=m\times c\times \Delta T$

q = heat released by the metal = -312.48 J

m = mass of metal = 43.1 g

c = specific heat of metal = ?

$\Delta T$ = change in temperature = $(T_{final}-T_{initial})=23.2-100=76.8^oC$

Now put all the given values in the above formula, we get:

$-312.48J=43.1g\times c\times 76.8^oC$

$c=0.0944J/g^oC$

Thus, the specific heat of the metal is $0.0944J/g^oC$

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