The question is incomplete, the complete question is;
A spherical steel ball bearing has a diameter of 2.540cm at 25.00∘C.
(a) What is its diameter when it's temperature is raised to
100∘C?
(b) What temperature change is required to increase its volume by
1.000% ?
Answer:
a) 2.542 cm
b) 303.03°C
Explanation:
Given;
Diameter of the ball= 2.540cm
Initial temperature= 25.0°C
Final temperature= 100.0°C
Percentage increase in volume = 1.000%
Temperature coefficient of expansion for steel =11.0×10^−6/∘C
d2= d1[1 + α(T2-T1)]
d2= 2.540[1 + 11.0×10^−6(100-25)]
d2= 2.540[1 + 8.25×10^-4]
d2= 2.542 cm
From;
%V ×1/100 = V ×3α ×∆T/ V
Substituting values;
1.000 ×1/100= 3× 11.0×10^−6 × ∆T
∆T= 0.01/3× 11.0×10^−6
∆T= 303.03°C
