Correct answers A)583N B)12.8miles A baseball (mass = 145 gm. surface area = 28.
ID: 2307021 • Letter: C
Question
Correct answers A)583N B)12.8miles A baseball (mass = 145 gm. surface area = 28.26 in^2) with a specific heat equal to 1.5 kJ/(kg K) is initially at a temperature of 75 degree F with the ambient at 100 degree F. It is accelerated to 90 miles/hr from a standstill over a time of 10 ms by a pitcher. Given that the ambient convective heat transfer coefficient is 20 W/(m^2 K) solve for: The force it takes to accelerate the ball in |N| The distance over which it takes the ball to reach 90degree F in [miles] Assume that the ball travels only in the x-direction with negligible drag and gravity forces influencing its flight. You will have to make an assumption for the average ball temperature over this distance to solve the problem. Show all calculations and unit conversions otherwise no points will be given.Explanation / Answer
mass m = 145 gm
intial vel = 0
final vel. = 90 miles/hr = 90*1.61/3.6 = 40.25 m/s
duration = 10 ms
accelration = 40.25/10e-3 = 4.025 e+3 m/s2
Force = ma = 145 e-3 * 4.025e+3 = 583.625 N
need to be applied to accelerate the ball
heat transfer co-efficent h = 20 w/m2 –k
ball temp = 750 F = 23.890 C = 296.89 K
ambient temp = 1000 F = 37.780 C = 310.78 K
surface area A = 28.26in2 = 0.0182 sq.m
heat transferred q = hA(Ta – Tb )
= 20*0.0182*(310.78-296.89)
= 5.05596 J/s
specific heat of the ball = 1.5kJ /kg-k
mass of the ball = 145 gms
heat required to raise the temperature to 900 F
= 0.145*1.5e+3*(15/1.8) = 1812 J
duration of heating = 1812/5.05 = 358 s
distance travled s = at2/2 = 4.025 e+3 *3582 /2
= 2.59e+8 m
= 2.59e+5 km
= 1.62 e+5 miles
The answer does not match with the expected answer. looks like something wrong with the given data like specific heat or heat transfer co-efficient or surface area.