Question

A 1.5 kg solid brass sphere is initially at a temperature of 15.0°C, the diameter of the brass sphere at this initial temperature is 6.50 cm.

(a) If the volume of the sphere is to increase by 2.50%, what is the final temperature of the brass sphere?

(b) If the sphere is heated to 225°C. What is the change in the radius of the sphere?

Answer 1

Given

The metal is brass

The coefficient of linear expansion for brass       α = 19 x 10^-6 m^Oc

The mass of the sphere                                                                m = 1.5 kg

The diameter of the sphere at initial condition d_o = 6.5 x 10^-2 m

The initial temperature                                                 T_o = 15.0^oC

He final temperature                                                     T = ?

Fourmula

ΔV/V_o = 3α (T-T_o)

V = V₀ [1+ 3α(T-T₀)]

Solution

The radius of the sphere r_o = d_o /2 = 6.5 x 10^-2 / 2 = 3.25 x 10^-2 m

The volume of the sphere at initial condition V_o = 4πr_o³/3

V_o = 4 x 3.14 x (3.25 x 10^-2)³ /3

V_o = 143.72 x 10^-6 m³

a)

the change in volume ΔV = 2.5% of V_o

ΔV = 143.72 x 10^-6 x 2.5/100

ΔV = 3.593 x 10^-6 m³

ΔV/V_o = 3α (T-T_o)

3.593/143.72 = 3 x 19 x 10^-6 (T - 15)

0.025 = 57 x 10^-6 (T - 15)

T – 15 = 438.6

T = 453.6^oC

b)

V = V₀ [1+ 3α(T-T₀)]

V = 143.72 x 10^-6[ 1+ 3 x 19 x 10^-6(225-15)]

V = 145.44 x 10^-6 m³

4πr³ /3 = 145.44 x 10^-6 m³

r³ = 3 x 145.44/4π

r = 3.26 x 10^-2 m