Question

A soccer goal has posts made of aluminum. The US Women's National Team is playing against Canada and the temperature outside is a very cold 0 degree C. (Assume the temperature of the air equals the temperature of the aluminum.) The goal opening has vertical posts 2.44 m tall and a horizontal cross-bar 7.32 m wide.

a-What is the area of the goal opening?

b-Having lost to the United States, Canada demands a rematch later in the year. The temperature outside is a very high 37.0 degree C. If the mass of the aluminum in the goal is 4280 kg, how much heat energy was absorbed by the aluminum?

c-How much longer are the posts and how much longer is the cross bar?

Answer 1

part a:

area of goal opening=length*width=2.44*7.32=17.8608 m^2

part b:

heat energy abosrbed=mass*specific heat of aluminium*temperature difference

=(4280 kg)*(900 J/kg.K)*(37-0)

=1.42524*10^8 Joules

part c:

let at 20 degree celcius the length be L.

coefficient of length expansion of aluminium at 20 degree celcius is a=23.1*10^(-6) /K

then length at 0 degree celcius=L*(1+a*(0-20))=L*(1-20*a)

this length is known to us

let us denote this as L1.

then L1=L*(1-20*a)

==>L=L1/(1-20*a)

length at 37 degree celcius=L*(1+a*(37-20))=L*(1+17*a)=L1*(1+17*a)/(1-20*a)

for the posts, height=L1=2.44 m

then new height=2.44*(1+17*a)/(1-20*a)=2.442086 m

for the cross bar, length is L1=7.32 m

then new length at 37 degree celclis=7.32*(1+17*a)/(1-20*a)=7.32626 m