A square plate has an area of 29.00 cm2 at 20.0C. It will be used in a low temperature experiment at T = 10.0K

linear-expansion
low-temperature

#1

A square plate has an area of 29.00 cm2 at 20.0C. It will be used in a low temperature experiment at T = 10.0K where it must have an area of 28.00 cm2. What area must be removed form the plate at 20.0C for it to have the correct area at 10.0 K? (The coefficient of linear expansion is 10  106 (C)1.)

Answer:

basically linear expansion is given by formula
linear area expansion is approx.= Area (initial)c(change in temp)
where c is coeff of linear area expansion expansion
And
final area expression as A(final)=A(initial)+change in area
A(final)=A(initial)(1+c(Δ.T))

just substitute A(final) =28 cm^2
A(initial)=x(unknown)
coeff of linear expansion is given so double it up for area expansion coeff
enter "c’=20*10^-6
chnge in temp would be = for this convert 20 degree celsius to kelvin and take change w.r.t to 10 k
change in temp would be =22.8

using above information calculate A(initial)
report answer as =29 cm^2 - A(initial)
report answer is your answer…

your welcome