Question

A nail driven into a board increases in temperature. If we assume that 60 % of the kinetic energy delivered by a 1.80-kg hammer with a speed of 7.80 m/s is transformed into heat that flows into the nail and does not flow out, what is the temperature increase of an 8.00-g aluminum nail after it is struck ten times?

Answer 1

Answer:

$\Delta T=45.63^{\circ}C$

Explanation:

Given:

• mass of hammer, $m=1.8\ kg$

• speed of the hammer, $v=7.8\ m.s^{-1}$

• part of kinetic energy of hammer getting transformed into heat, $F=60\%$

• mass of nail, $m'=8\ g=0.008\ kg$

• we've specific heat of aluminium, $c=900\ J.kg^{-1}.^{\circ}C^{-1}$

Total kinetic energy given to the nail:

$KE=10\times \frac{1}{2} \times m.v^2$ (Since the aluminium nail is struck 10 times)

The heat that flows to the nail is 0.6 times of the total kinetic energy:

$Q=0.6\times 10\times 0.5\times1.8\times 7.8^2$

$Q=328.536\ J$

From the equation of heat:

$Q=m'c.\Delta T$

$328.536=0.008\times 900\times \Delta T$

$\Delta T=45.63^{\circ}C$ is the increase in the temperature of the nail.

Answer 2

The increase in the temperature will be "45.63°C".

According to the question,

Mass,

• $m = 1.8 \ kg$

Speed,

• $v = 7.8 \ m.s^{-1}$

K.E getting transformed,

• $F = 60$%

Mass of nail,

• $m' = 8 \ g$

or,

             $= 0.008 \ kg$

Specific heat,

• $c = 900 \ J.kg^{-1}.^{\circ} C^{-1}$

Now,

The total K.E given to the nail will be:

= $10\times \frac{1}{2} mv^\\2$

The heat flows will be:

$Q = 0.6\times 10\times 0.5\times 1.8\times 7.8^2$

   $= 328.536 \ J$

By using the equation of heat, we get

→         $Q = m'c.\Delta T$

→ $328.536=0.008\times 900\times \Delta T$

→       $\Delta T = 45.63^{\circ} C$

Thus the above response is correct.  

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