Specific heat capacity (Calorimetry)

[Lunar Guy]

[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 [tex](c_{p} = [/tex] 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

[tex]c_{p,x}m_{x}T_{x}[/tex] = [tex]c_{p,y} m_{y} T_{y}[/tex]

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...

[tex]c_{p,x}m_{x}T_{x}[/tex] = [tex]c_{p,y} m_{y} T_{y}[/tex]

 

[Lunar Guy]

Bump. Can anyone help me out?

 

[blochwave]

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)

 

[Lunar Guy]

[tex]\Delta[/tex]PE + [tex]\Delta[/tex]KE + [tex]\Delta[/tex]U = 0?

 

[blochwave]

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)

 

[Lunar Guy]

Thanks. I got it. It's 120 degrees Celsius, right?