Question

The 1.30-kg head of a hammer has a speed of 7.3 m/s just before it strikes a nail and is brought to rest Estimate the temperature rise of a 14-g iron nail generated by 8.0 such hammer blows done in quick succession. Assume the nail absorbs all the energy. The specific heat of iron is 450 J/kg⋅C∘.

Answer 1

Answer:

The rise in temperature is $43.98^{\circ}C$

Solution:

As per the question:

Mass of hammer, M = 1.30 kg

Speed of hammer, v = 7.3 m/s

Mass of iron, $m_{i} = 14\ g$

No. of blows, n = 8

Specific heat of iron, $C_{i} = 450\ J/kg.^{\circ}C = 0.45\ J/g^{\circ}C$

Now,

To calculate the temperature rise:

Transfer of energy in a blow = Change in the Kinetic energy

$\Delta KE = \frac{1}{2}Mv^{2} = \frac{1}{2}\times 1.30\times 7.3^{2} = 34.64\ J$

For 8 such blows:

$\Delta KE = n\Delta KE = 8\times 34.64\ = 277.12 J$

Now, we know that:

$Q = m_{i}C_{i}\Delta T$

$\Delta T= \frac{\Delta KE}{m_{i}C_{i}}$

$\Delta T= \frac{277.12}{14\times 0.45} = 43.98^{\circ}C$