Answer:
0.921 J/g degrees C
Explanation:
Recall that the First Law of Thermodynamics demands that the total internal energy of an isolated system must remain constant. Any amount of energy lost by the brakes must be gained by the tires (in the form of heat in this situation). Therefore, heat given off by the brakes = −heat taken in by tires, or:
−qbrakes=qtires
The equation used to calculate the quantity of heat energy exchanged in this process is:
−qbrakes=−cbrakes mbrakes ΔTbrakes=ctires mtires ΔTtires=qtires
First we must convert the mass of the tires and the brakes from kg to g.
massbrakes=90.7 kg×1,000. g1 kg=9.07×104 g
masstires=123 kg×1,000. g1 kg=1.23×105 g
Next, substitute in known values and rearrange to solve for ctires. Note that the final temperature for both the tires and the brakes is 172∘C, the initial temperature of the brakes is 312∘C and the initial temperature of the tires is 15∘C.
−(1.400Jg∘C)(9.07×104 g)(172∘C−312∘C)=(ctires)(1.23×105 g)(172∘C−15∘C)
ctires=−(1.400 Jg∘C)(9.07×104 g)(−140∘C)(1.23×105 g)(157∘C)=17,777,200 J19311000 g∘C=0.9206Jg∘C
The answer should have three significant figures, so round to 0.921Jg∘C.
of CO2