Question

A man (mass 114 kg) runs with speed 5.7 m/s along level ground. His boots (which were already included in his total mass) have a mass of 640 g, and a specific heat of 1200 J/(kg K). His boots are initially at a temperture of 12°C.


He slides to a halt converting all of his mechanical energy into thermal energy. Assuming all of the thermal energy is absorbed by his boots, what is the temperature of his boots after sliding to a stop?

Answer 1

The mechanical energy of the man is its kinetic energy.

Now, the kinetic energy of the man is

K.E = ½ mv²

Here, mass of the man is m and speed of the man is v.

The initial temperature of the boots is

T_i = 12 °C

    = 12 + 273 K

    = 285 K

Let T_f be the final temperature of the boot which is the temperature of the boots after sliding to a stop.

The amount of thermal energy required to increases the temperature of the boot is

H = m_Bc_B (T_f - T_i)

Here, mass of the boots is m_B and specific heat of the boot is c_B.

The mechanical energy of the man is converted into the thermal energy which is absorbed by the boot.

From the law of conservation of energy,

Substitute 114 kg for m, 5.7 m/s for v, 640 g for m_B, 1200 J/(kg K) for c_B and 285 K for T_i in the above equation,

Rounding off to three significant figures, the temperature of his boots after sliding to a stop is 287 K or 14.4 °C.