Question

A lead bullet (specific heat=\[0.032\,{\text{cal/g}}^\circ {\text{C}}\]) is completely stopped when it strikes a target with a velocity \[300\,{\text{m/s}}\]. The heat generated is equally shared by the bullet and the target. The rise in temperature of the bullet will be
A. \[16.7^\circ {\text{C}}\]
B. \[1.67^\circ {\text{C}}\]
C. \[167.4^\circ {\text{C}}\]
D. \[267.4^\circ {\text{C}}\]

Answer 1

Hint: Use the formula for kinetic energy of an object. Also use the formula for the heat exchanged by the substance. Use the law of conservation of angular momentum. According to this law, the half of the initial kinetic energy of the lead bullet is equal to the heat shared by the lead bullet and the target.

Formulae used:
The kinetic energy \[K\] of an object is
\[K = \dfrac{1}{2}m{v^2}\] …… (1)
Here, \[m\] is the mass of the object and \[v\] is the velocity of the object.
The heat \[Q\] exchanged by the substance is
\[Q = mc\Delta T\] …… (2)
Here, \[m\] is the mass of the substance, \[c\] is specific heat of the substance and \[\Delta T\] is a change in temperature of the substance.

Complete step by step answer:
We have given that the specific heat of the lead bullet is \[0.032\,{\text{cal/g}}^\circ {\text{C}}\].
\[c = 0.032\,{\text{cal/g}}^\circ {\text{C}}\]
The velocity of the bullet is \[300\,{\text{m/s}}\].
\[v = 300\,{\text{m/s}}\]
We have asked to calculate the rise in temperature of the bullet.

Let us first calculate the kinetic energy of the lead bullet.Let us consider the mass of the lead bullet is \[m\].Substitute \[300\,{\text{m/s}}\] for \[v\] in equation (1).
\[K = \dfrac{1}{2}m{\left( {300\,{\text{m/s}}} \right)^2}\]
\[ \Rightarrow K = 45000m\]
Hence, the kinetic energy of the lead bullet is \[45000m\].

We can now calculate the heat gained by the lead bullet when it strikes the target.Convert the unit of specific heat of the lead bullet to the SI system of units.
\[c = \left( {0.032\,{\text{cal/g}}^\circ {\text{C}}} \right)\left( {\dfrac{{4.2 \times {{10}^3}\,{\text{J}}}}{{1\,{\text{cal}}}}} \right)\]
\[ \Rightarrow c = 134.4\,{\text{J/g}}^\circ {\text{C}}\]
Hence, the specific heat of the lead bullet is \[134.4\,{\text{J/g}}^\circ {\text{C}}\].
Substitute \[134.4\,{\text{J/g}}^\circ {\text{C}}\] for \[c\] in equation (2).
\[Q = m\left( {134.4\,{\text{J/g}}^\circ {\text{C}}} \right)\Delta T\]

According to the law of conservation of energy, half of the initial kinetic energy of the lead bullet is equal to the heat exchanged by the bullet with the target as the heat is shared equally by the bullet and the lead.
\[\dfrac{K}{2} = Q\]
Substitute \[45000m\] for \[K\] and \[m\left( {134.4\,{\text{J/g}}^\circ {\text{C}}} \right)\Delta T\] for \[Q\] in the above equation.
\[\dfrac{{45000m}}{2} = m\left( {134.4\,{\text{J/g}}^\circ {\text{C}}} \right)\Delta T\]
\[ \Rightarrow \dfrac{{45000}}{2} = \left( {134.4\,{\text{J/g}}^\circ {\text{C}}} \right)\Delta T\]
\[ \Rightarrow \Delta T = \dfrac{{45000}}{{2 \times 134.4\,{\text{J/g}}^\circ {\text{C}}}}\]
\[ \therefore \Delta T = 167.4^\circ {\text{C}}\]
Hence, the rise in temperature of the bullet is \[167.4^\circ {\text{C}}\].

Hence, the correct option is C.

Note: The students should not forget to convert the unit of specific heat of the lead bullet to the SI system of units as all the physical quantities used in the formulae are in the SI system of units. The students should also not forget to use half of the kinetic energy of the bullet equal to the heat shared by the bullet and the target.