Question

Upon landing, the 90.7 kg carbon fiber brakes of an airliner heat up 312∘C, producing heat. As the brakes start to cool back to their initial temperature, the heat is absorbed by the 123 kg rubber tires. Assuming that all of the heat is transferred from the brakes to the tires, what is the specific heat of the tires if their temperature rises 172∘C?
Round your answer to two decimal places.
Use 1.400Jg∘C for the specific heat of carbon fiber.

Answer 1

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

Answer 2

Answer: The specific heat of the tires is $1.87 Jg^0C$

Explanation: 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.

$Q= m\times c\times \Delta T$

Q= heat gained  or lost

m= mass of the substance

c = heat capacity of substance    

$\Delta T={\text{Change in temperature}}$ 

Heat lost by the carbon fiber brakes = Heat gained by the rubber tires

$90700g\times 1.400Jg^0C\times 312^0C = 123000g\times c\times 172^0C$

$c= 1.87Jg^0C$

Therefore, the specific heat of rubber tires to two decimal places is $1.87Jg^0C$