On a beautiful baseball day with air temperature of 20 0 C, a pitcher throws a 0

Question

On a beautiful baseball day with air temperature of 200 C, a pitcher throws a 0.142 kg baseball at 49 m/s towards home plate. As it travels 19.4 m, the ball slows down to a speed of 36 m/s because of air resistance. Calculate the change in the temperature of the air through which the ball passes. For air M= 28.9 g/mol and Cp =7R/2. Hint: Assume the change in temperature happens only for a cylinder of air 19.4 m in length and radius of 3.7 cm. Assume that the energy loss of the ball due to its change in KE and is all absorbed by the air.

Explanation / Answer

the volume of air column is

pi*r^2 *d = pi*0.037^2*19.4 = 0.08344 m^3 = 83.44 L

the density of air can be found by ideal gas law

PV = nRT = m/M RT

P = /M *RT

= 1.013*10^5*0.0289/(8.31*293.15) =1.202 g/L

m = 1.202*83.44 =   100.3 g

n = m/M = 100.3/28.9 = 3.47 mol

the lost of Ke of the ball

1/2 *0.142*(49^2-36^2) =78.46 J

78.46 J = 3.47 *7R/2 delta T

delta T = 0.78 K

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