Question

The 1.19 kg head of a hammer has a speed of 8.4 m/s just before it strikes a nail (Fig. 14-13) and is brought to rest. Estimate the temperature rise of a 13 g iron nail generated by twenty such hammer blows done in quick succession. Assume the nail absorbs all the energy.

Answer 1

Answer:

The temperature rise is $143.9\ K$.

Explanation:

Given the mass of the hammer $(m)$ is 1.19 kg.

And the speed of the hammer $(v)$ is 8.4 m/s.

We need to find the temperature change of iron nail having 20 hammer blows.

First, we will find the kinetic energy of the each blow of hammer.

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

$K.E=\frac{1}{2}\times 1.19\times 8.4^2=41.99\ J$

For 20 such blows, the kinetic energy will be

$20\times K.E=20\times 41.99=839.8\ J$

Let us assume that the nail absorbs all the kinetic energy. So, this 839.8 J will be converted into heat energy $(Q)$.

Now,

$Q=839.8\ J$

Also, we know that

$Q=mc\Delta_T$

Where,

$m$ is the mass of the nail, which is $13\ g=\frac{13}{1000}=0.013\ kg$

$c$ is the specific heat capacity of iron, which is 449 J/kg.K

$\Delta_T$ is the change in temperature.

Plug these value we get,

$Q=mc\Delta_T\\839.8\ J=(0.013\ kg)(449\ J/kg.K)\times\Delta_T\\\Delta_T=\frac{839.8}{0.013\times 449}\\\\\Delta_T=143.9\ K$

We can see the temperature rise is 143.9 K.