Question

A nail driven into a board increases in temperature.

Answer 1

Answer:

ΔT = 40.91 °C

Explanation:

First we find the kinetic energy of one hit to the nail:

K.E = (1/2)mv²

where,

K.E = Kinetic energy = ?

m = mass of hammer = 1.6 kg

v = speed of hammer = 7.7 m/s

Therefore,

K.E = (1/2)(1.6 kg)(7.7 m/s)²

K.E = 47.432 J

Now, for 10 hits:

K.E = (10)(47.432 J)

K.E = 474.32 J

Now, we calculate the heat energy transferred (Q) to the nail. As, it is the 59% of K.E. Therefore,

Q = (0.59)K.E

Q = (0.59)(474.32 J)

Q = 279.84 J

The change in energy of nail is given as:

Q = mCΔT

where,

m = mass of nail = 7.6 g = 0.0076 kg

C = specific heat capacity of aluminum = 900 J/kg.°C

ΔT = Increase in temperature = ?

Therefore,

279.84 J = (0.0076 kg)(900 J/kg.°C)ΔT

ΔT = (279.84 J)/(6.84 J/°C)

ΔT = 40.91 °C