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Lemur [1.5K]
3 years ago
12

The temperature at the surface of the Sun is approximately 5,300 K, and the temperature at the surface of the Earth is approxima

tely 293 K. What entropy change of the Universe occurs when 6.00 103 J of energy is transferred by radiation from the Sun to the Earth?
Physics
1 answer:
N76 [4]3 years ago
4 0

Answer:

The entropy change of the Universe that occurs is 19.346 J/K

Explanation:

Given;

temperature of the sun, T_s = 5,300 K

temperature of the Earth, T_E = 293 K

radiation energy transferred by the sun to the earth, E = 6000 J

The sun loses Q of heat and therefore decreases its entropy by the amount

\delta S_{sun} = \frac{-Q}{T_s}

The earth gains Q of heat and therefore increases its entropy by the amount

\delta S_{Earth} = \frac{-Q}{T_E}

The total entropy change is:

\delta S_{Earth} + \delta S_{sun} = \frac{Q}{T_E} -\frac{Q}{T_S} \\\\                                                      = Q(\frac{1}{T_E} -\frac{1}{T_S} )\\\\= 6000(\frac{1}{293} -\frac{1}{5300} )\\\\=6000(0.0032243)\\\\= 19.346 \ J/K

Therefore, the entropy change of the Universe that occurs is 19.346 J/K

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Answer:

a) 10.51 J

b) 3.48 m/s

Explanation:

Given data :

mass of train ( M ) = 2.2 kg

Given initial velocity ( u ) = 1.6 m/s

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F = mx + b  ------ ( 1 )

m = slope  = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m

x = distance travelled on the x axis by the train = 7.5 m

F = force experienced by the train = 2.8 N

x = 0

∴ b = 2.8

hence equation 1 can be written as

F = ( -0.373) x + 2.8   ----- ( 2 )

hence to determine the work done by the force

W   = \int\limits^7_0 { ( -0.373) x + 2.8  )} \, dx     Note:  the limits are actually 7.5 and 0

∴ W ( work done ) = -10.49 + 21 = 10.51 J

<u>b) calculate the speed of the train at the end of its journey</u>

we will apply the work energy theorem

W = 1/2 m*v^2  -  1/2 m*u^2

∴ V^2 = 2 / M ( W + 1/2 M*u^2 )  ( input values into equation )

 V^2 = 12.11

hence V = 3.48 m/s

6 0
3 years ago
A 0.877 mol sample of N2(g) initially at 298 K and 1.00 atm is held at constant volume while enough heat is applied to raise the
MissTica

Answer : The value of q\text{ and }\Delta U is 286.2 J and 286.2 J respectively.

Explanation : Given,

Moles of sample = 0.877 mol

Change in temperature = 15.7 K

First we have to calculate the heat absorbed by the system.

Formula used :

q=n\times c_v\times \Delta T

where,

q = heat absorbed by the system = ?

n = moles of sample = 0.877 mol

\Delta T = Change in temperature = 15.7 K

c_v = heat capacity at constant volume of N_2 (diatomic molecule) = \frac{5}{2}R

R = gas constant = 8.314 J/mol.K

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

q=0.877mol\times \frac{5}{2}\times 8.314J/mol.K\times 15.7K

q=286.2J

Now we have to calculate the change in internal energy of the system.

\Delta U=q+w

As we know that, work done is zero at constant volume. So,

\Delta U=q=286.2J

Therefore, the value of q\text{ and }\Delta U is 286.2 J and 286.2 J respectively.

8 0
4 years ago
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The answer to this statement is true!
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Tems11 [23]

Answer:

8 seconds

Explanation:

power (P) is defined as the rate at which work is done.

power is measured in Watts (W) , when the work done is measured in Joules (J) and time in seconds

by the definition of power,

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3 0
4 years ago
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Answer:

F = 39.36 N

Explanation:

given,

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using impulse formula

J = change in momentum

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F = 39.36 N

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