Answer:
Given,
Mass of carbon fiber brakes = 90.7 kg = 90.7 x 1000 = 90700 g
Mass of the rubber tires = 123 kg = 123 x 1000 = 123000
Specific heat of carbon fiber = 1.400Jg∘C
As all the heat is transferred from the carbon fiber brakes to the rubber tires, the heat lost by the carbon fiber brakes is equal to the heat gained by the rubber tires.
So the temperature change of carbon fiber brakes, delta T = 172∘C
And the temperature change of rubber tires, delta T = 172∘C
The formula used for heat is,
Q = m s delta T
where Q is heat, m is mass, s is specific heat, and delta T is the temperature change
We know,
Heat lost by the carbon fiber brakes = Heat gained by the rubber tires
90700 g x 1.400Jg∘C x 172∘C = 123000 g x s x 172∘C
or, 90700 g x 1.400Jg∘C = 123000 g x s
or, s = (90700 g x 1.400Jg∘C) / 123000 g
or, s = 1.03236
Therefore, the specific heat of rubber tires is 1.03236 Jg∘C
Explanation: