A mass M subway train initially traveling at speed v slows to a stop in a station and then stays there long enough for its brakes to cool. The station's volume is V and the air in the station has a density rho_air and specific heat C_air.
Assuming all the work done by the brakes in stopping the train is transferred as heat uniformly to all the air in the station, what is the expression for the change in the air temperature in the station? Make sure the quantities you enter match the ones given in the problem exactly.

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okpalawalter8 okpalawalter8 · Answer 1 · 2020-06-03T02:17:11+02:00

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Answer:

The expression for the change in the air temperature is [tex]\Delta T = \frac{Mv^2}{2 \rho_{air} c_{air}* V}[/tex]

Explanation:

From the question we are told that

The mass of the train is M

The speed of the train is v

The volume of the station is V

The density of air in the station is [tex]\rho_{air}[/tex]

The specific heat of air is [tex]c_{air}[/tex]

The workdone by the break can be mathematically represented as

[tex]W =\Delta KE = \frac{1}{2} Mv^2[/tex]

Now this is equivalent to the heat transferred to air in the station

Now the heat capacity of the air in the station is mathematically represented as

[tex]Q = \rho_{air} * m_{air} * c_{air} (\Delta T)[/tex]

Now Since this is equivalent to the workdone by the breaks we have that

[tex]\frac{1}{2} Mv^2 = m_{air} * c_{air} (\Delta T)[/tex]

=> [tex]\Delta T = \frac{Mv^2}{2 \rho_{air} c_{air}* V}[/tex]