A hollow sphere is made from copper (Cu). The inner and outer diameters are 10 a

Question

A hollow sphere is made from copper (Cu). The inner and outer diameters are 10 and 8 cm, respectively.
a) Compute the mass m of the copper sphere (not counting the air in the cavity).
b) Compute the number n of moles of copper.
c) Compute the number N of copper atoms.

I need help with number 2 please

#2The copper sphere described above was at 20 °C. You immerse the sphere in boiling water.
a) Compute the fractional and absolute change of the inner diameter of the sphere (i.e. d/d and d).
Hint: Consult a book or the internet to find any required physical properties of copper (if they are not
provided above).
b) Compute the fractional and absolute change of the inner cavity (i.e. the volume of gas) in the sphere.
c) By how much does the number of copper atoms change?

Explanation / Answer

a)

volume of the copper V = (4/3)*pi*(R^2-r^2)

R = 10/2 = 5 cm = 0.05 m

r = 8/2 = 4 cm = 0.04 m

V = (4/3)*pi*(0.05^3-0.04^3)

V = 0.000256 m^3

mass of copper m = D*V = 8940*0.000256 = 2.29 kg

===========

(b)

number of moles = m/M

M = molar mass

number of moles n = 2.29/0.063546 = 36 moles

===================

part (c)

Number of atoms N = n*A

A = avagadro number

N = 36*6.023*10^23

N = 2.17*10^25 <<<-------------ANSWER

============================

#2)

fractional change dr/r = alpha*dT = 17*10^-6*(100-20) = 0.00136

obsolute change dr = r*alpha*dT = 0.00136*0.04 = 5.44*10^-6 m

(b)

fractional change = 3*dr/r = 3*0.00136 = 0.00408

absolute change = (4/3)*pi*(r1^3-r^3)

r1 = r + dr = 0.04 + 5.44*10^-6 = 0.04000544 m

absolute change = (4/3)*pi*(0.04000544^3-0.04^3) = 1.1*10^-7 m^3

c)

number of atoms remain same

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