Hot wheels

Classical Mechanics Level 2

A 750 kg car moving at 24 m/s brakes to a stop. The brakes contain about 15 kg of iron, which absorbs the energy. What is the increase in temperature of the brakes?

Hint: The specific heat capacity of iron equals 450 J/kg K.

32° C 256° C 512° C 128° C 64° C

Markus Michelmann by Markus Michelmann , 35, Germany

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Markus Michelmann
Sep 10, 2017

The initial kinetic energy of the car reads

E kin = 1 2 m car v 2 = 1 2 ⋅ 750 kg ⋅ ( 24 m s 2 ) 2 = 216 kJ E_\text{kin} = \frac{1}{2} m_\text{car} v^2 = \frac{1}{2} \cdot 750 \,\text{kg} \cdot (24 \,\frac{\text{m}}{\text{s}^2})^2 = 216 \, \text{kJ}

This energy is converted into heat

E heat = m brake c iron Δ T = ! E kin E_\text{heat} = m_\text{brake} c_\text{iron} \Delta T \stackrel{!}{=} E_\text{kin}

⇒ Δ T = E kin m brake c iron = 216 kJ 15 kg ⋅ 450 J kg K = 32 K \Rightarrow \quad \Delta T = \frac{E_\text{kin}}{m_\text{brake} c_\text{iron}} = \frac{ 216 \, \text{kJ}}{15\, \text{kg} \, \cdot 450 \, \frac{\text{J}}{\text{kg}\,\text{K}} } = 32 \, \text{K}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...