A2 kg subway train initially traveling at 18.5 m/s slows to a stop in a station and then stays there long enough for its brakes to cool. the station's dimensions are 65.0 m long by 20.0 m wide by 12.0 m high. assuming all the work done by the brakes in stopping the train is transferred as heat uniformly to all the air in the station, by how much does the air temperature in the station rise? take the density of the air to be 1.20 kg/m3 and its specific heat to be 1020 j/(kg*k).

rise the air temperature is 0.179241 K

Explanation:

Given data

mass = 20000 kg

velocity  = 18.5 m/s

long = 65 m

wide = 20 m

height = 12 m

density of the air = 1.20 kg/ m³

specific heat = 1020 J/(kg*K)

to find out

how much does the air temperature in the station rise

solution

we know here Energy lost by the train that is calculated by  

loss in the kinetic energy that is = 1/2 m v²

loss in the kinetic energy = 0.5 × 20000 ×18.5²

loss in the kinetic energy  is 3422500 J

and

this energy is used here to rise the air temperature that is KE / ( specific hat × mass )

so here  

air volume = 65 ×20×12

air volume = 15600 m³

air mass  =  ρ × V = 1.2 × 15600

air mass = 18720 kg

so

rise the air temperature = 3422500 / ( 1020 × 18720)

rise the air temperature is 0.179241 K

i think the answer is tomato juice? not entirely sure though

at right angles to each other