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A $20 \,g$ bullet whose specific heat is $5000 \,J kg ^{\circ} C$ and moving at $2000 \,m / s$ plunges into a $1.0 \,kg$ block of wax whose specific heat is $3000 \,J kg ^{\circ} C$. Both bullet and wax are at $25^{\circ} C$ and assume that $(i)$ the bullet comes to rest in the wax and $(ii)$ all its kinetic energy goes into heating the wax. Thermal temperature of the wax $\left(\right.$ in $\left.^{\circ} C \right)$ is close to
$28.1$
$31.5$
$37.9$
$42.1$
Solution
(c)
As kinetic energy of bullet is used up in heating and melting the wax.
By energy conservation, we have
$\frac{1}{2} m_b v_b^2=m_w c_w\left(\Delta T_w\right)+m_b c_b\left(\Delta T_b\right)$
As both bullet and wax initially are at same temperature $\left(T_i=25^{\circ} C \right)$.
So, $\Delta T_w=\Delta T_b=\Delta T$ (say)
Then, $\frac{1}{2} m_b v_b^2=\left(m_w c_w+m_b c_b\right) \Delta T$
or $\Delta T=\frac{m_b v_b^2}{2\left(m_w c_w+m_b c_b\right)}$
Substituting values in above equation, we get
$\Delta T=\frac{20 \times 10^{-3} \times(2000)^2}{2\left(20 \times 10^{-3} \times 5000+1 \times 3000\right)}$
$\Rightarrow \Delta T=\frac{400}{31}=12.9 \Rightarrow T_f-T_i=12.9$
$\text { or }T_f=25+12.9=37.9^{\circ} C$
