Answer to Question #158193 in Molecular Physics | Thermodynamics for Harshita

Question

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A 1200 kg car moving at 12m/s applies it's break and stops . If the brakes are made of 5 kg of steel ( C = 450 JK ^-1 Kg^-1) each for 4 brakes ,by how much does the temperature of the metal rise ?

Answer 1

First, let's write the kinetic energy lost due to braking:

$E = \frac{1}{2}mv^2$

Now, if we suppose that all this energy is converted to heat the breaks, we have:

$E = C\cdot m_{brake}\cdot4\cdot \Delta T$

By using the former expression:

$\frac{1}{2}m_{car}v^2 = C \cdot m_{brake}\cdot 4 \cdot \Delta T$

$\Delta T = \frac{m_{car}v^2}{8Cm_{brake}}=\frac{1200\cdot 144}{8\cdot450\cdot 5} = 9.6 ^{\circ}C (=9.6K)$