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Question: A lead bullet (specific heat=\[0.032\,{\text{cal/g}}^\circ {\text{C}}\]) is completely stopped when ...

A lead bullet (specific heat=0.032cal/gC0.032\,{\text{cal/g}}^\circ {\text{C}}) is completely stopped when it strikes a target with a velocity 300m/s300\,{\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. 16.7C16.7^\circ {\text{C}}
B. 1.67C1.67^\circ {\text{C}}
C. 167.4C167.4^\circ {\text{C}}
D. 267.4C267.4^\circ {\text{C}}

Explanation

Solution

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 KK of an object is
K=12mv2K = \dfrac{1}{2}m{v^2} …… (1)
Here, mm is the mass of the object and vv is the velocity of the object.
The heat QQ exchanged by the substance is
Q=mcΔTQ = mc\Delta T …… (2)
Here, mm is the mass of the substance, cc is specific heat of the substance and ΔT\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 0.032cal/gC0.032\,{\text{cal/g}}^\circ {\text{C}}.
c=0.032cal/gCc = 0.032\,{\text{cal/g}}^\circ {\text{C}}
The velocity of the bullet is 300m/s300\,{\text{m/s}}.
v=300m/sv = 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 mm.Substitute 300m/s300\,{\text{m/s}} for vv in equation (1).
K=12m(300m/s)2K = \dfrac{1}{2}m{\left( {300\,{\text{m/s}}} \right)^2}
K=45000m\Rightarrow K = 45000m
Hence, the kinetic energy of the lead bullet is 45000m45000m.

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=(0.032cal/gC)(4.2×103J1cal)c = \left( {0.032\,{\text{cal/g}}^\circ {\text{C}}} \right)\left( {\dfrac{{4.2 \times {{10}^3}\,{\text{J}}}}{{1\,{\text{cal}}}}} \right)
c=134.4J/gC\Rightarrow c = 134.4\,{\text{J/g}}^\circ {\text{C}}
Hence, the specific heat of the lead bullet is 134.4J/gC134.4\,{\text{J/g}}^\circ {\text{C}}.
Substitute 134.4J/gC134.4\,{\text{J/g}}^\circ {\text{C}} for cc in equation (2).
Q=m(134.4J/gC)ΔTQ = 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.
K2=Q\dfrac{K}{2} = Q
Substitute 45000m45000m for KK and m(134.4J/gC)ΔTm\left( {134.4\,{\text{J/g}}^\circ {\text{C}}} \right)\Delta T for QQ in the above equation.
45000m2=m(134.4J/gC)ΔT\dfrac{{45000m}}{2} = m\left( {134.4\,{\text{J/g}}^\circ {\text{C}}} \right)\Delta T
450002=(134.4J/gC)ΔT\Rightarrow \dfrac{{45000}}{2} = \left( {134.4\,{\text{J/g}}^\circ {\text{C}}} \right)\Delta T
ΔT=450002×134.4J/gC\Rightarrow \Delta T = \dfrac{{45000}}{{2 \times 134.4\,{\text{J/g}}^\circ {\text{C}}}}
ΔT=167.4C\therefore \Delta T = 167.4^\circ {\text{C}}
Hence, the rise in temperature of the bullet is 167.4C167.4^\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.