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

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 = (1)/(2)Mv^(2) = (1)/(2)* 1.30* 7.3^(2) = 34.64\ J`

For 8 such blows:

`\Delta KE = n\Delta KE = 8* 34.64\ = 277.12 J`

Now, we know that:

`Q = m_(i)C_(i)\Delta T`

`\Delta T= (\Delta KE)/(m_(i)C_(i))`

`\Delta T= (277.12)/(14* 0.45) = 43.98^(\circ)C`