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(II)The 1.20-kg head of a hammer has a speed of 7.5 m/s just before it strikes a nail (Fig. 14–17) and is brought to rest. Estimate the temperature rise of a 14-g iron nail generated by eight such hammer blows done in quick succession. Assume the nail absorbs all the energy.

Short Answer

Expert verified

The rise in temperature is \(42.8\circ {\rm{C}}\).

Step by step solution

01

Given data

The mass of the head of the hammer is\(m = 1.20\;{\rm{kg}}\).

The mass of the iron nail is\(m' = 14\;{\rm{g}}\).

The speed is \(v = 7.5\;{\rm{m/s}}\).

02

Understanding kinetic energy

In this problem, the relation of kinetic energy and heat transfer will be used to determine the change in temperature.

03

Calculation of the rise in temperature

The relation to find the temperature is given by:

\(\begin{array}{c}KE = Q\\8 \times \frac{1}{2}m{v2} = m'c\Delta T\end{array}\)

Here, cis the specific heat,\(\Delta T\)is the change in temperature, Qis the heat transfer, and KE is the kinetic energy.

On plugging the values in the above relation, you get:

\(\begin{array}{c}8 \times \frac{1}{2}\left( {1.20\,{\rm{kg}}} \right){\left( {7.5\;{\rm{m/s}}} \right)2} = \left( {14\,{\rm{g}} \times \frac{{{{10}{ - 3}}\;{\rm{kg}}}}{{1\;{\rm{kg}}}}} \right)\left( {450\;{\rm{J/kg}} \cdot \circ {\rm{C}}} \right)\Delta T\\\Delta T = 42.8\circ {\rm{C}}\end{array}\)

Thus, \(\Delta T = 42.8\circ {\rm{C}}\) is the required rise in temperature.

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