Specific heat capacity (Calorimetry)

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Homework Help Overview

The problem involves the conversion of kinetic energy to internal energy in the context of an automobile's braking system. It specifically examines how the temperature of steel brake disks changes as a result of this energy conversion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the appropriate equations related to specific heat and energy transfer. There are attempts to clarify the relationship between kinetic energy and heat energy, as well as the distribution of energy among the brake disks.

Discussion Status

Some participants have provided guidance on the relevant equations and concepts, while others are exploring different interpretations of the problem. There is an ongoing exchange of ideas regarding the correct approach to the calculations involved.

Contextual Notes

The original poster expresses uncertainty about how to begin the problem, indicating a need for foundational understanding. There is also a mention of the assumption that all kinetic energy is converted to heat, which is a critical point of discussion.

Lunar Guy
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[SOLVED] Specific heat capacity (Calorimetry)

Homework Statement


When a driver brakes an automobile, friction between the brake disks and the brake pads converts part of the car's translational kinetic energy to internal energy. If a 1500 kg automobile traveling at 32 m/s comes to a halt after its brakes are applied, how much can the temperature rise in each of the four 3.5 kg steel brake disks? Assume the disks are made of iron (c_{p} = 448 J/kg•°C) and that all of the kinetic energy is distributed in equal parts to the internal energy of the brakes.


Homework Equations



c_{p,x}m_{x}T_{x} = c_{p,y} m_{y} T_{y}




The Attempt at a Solution


I really don't know where to start with this problem... A little push and I can do the rest, but...

c_{p,x}m_{x}T_{x} = c_{p,y} m_{y} T_{y}
 
Last edited:
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Bump. Can anyone help me out?
 
Well that's not really the right equation involving specific heat you're looking at

You want the one that involves energy. You know the kinetic energy of the car from 1/2*mv^2, and it's given that all that energy becomes heat

A 4th of that goes into each 3.5kg brake, so do you have a formula that links specific heat, mass, temperature, and heat? (yes you do)
 
Last edited:
\DeltaPE + \DeltaKE + \DeltaU = 0?
 
Well you could infer the equation from the units if you didn't know off hand(energy/mass*temp

c=Q/(m*dT) where dT is the change in temperature and Q is the amount of heat gained(which you find from assuming all the kinetic energy becomes heat)
 
Thanks. I got it. It's 120 degrees Celsius, right?
 

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