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A $20 \mathrm{gm}$ bullet whose specific heat is $5000 \mathrm{~J} /\left(\mathrm{kg}-{ }^{\circ} \mathrm{C}\right)$ and moving at $2000 \mathrm{~m} / \mathrm{s}$ plunges into a $1.0 \mathrm{~kg}$ block of wax whose specific heat is $3000 \mathrm{~J} /\left(\mathrm{kg}-{ }^{\circ} \mathrm{C}\right)$. Both bullet and wax are at $25^{\circ} \mathrm{C}$ 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 in ${ }^{\circ} \mathrm{C}$ is close to
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The correct answer is:
$37.9$
$\mathrm{M}_{\mathrm{B}}=20 \times 10^{-3} \mathrm{Kg}$
$\mathrm{C}_{\mathrm{B}}=5000 \mathrm{~J} / \mathrm{Kg}-{ }^{\circ} \mathrm{C}$
$\mathrm{V}=2000 \mathrm{M} / \mathrm{s}$
$\mathrm{M}_{\mathrm{w}}=1 \mathrm{Kg}$
$\mathrm{C}_{\mathrm{w}}=3000 \mathrm{~J} / \mathrm{Kg}-{ }^{\circ} \mathrm{C}$
$\mathrm{T}_{\mathrm{f}}=25^{\circ} \mathrm{C}=298 \mathrm{~K}$
$\frac{1}{2} \mathrm{MV}^{2}=\mathrm{M}_{\mathrm{w}} \mathrm{C}_{\mathrm{w}} \Delta \mathrm{T}_{\mathrm{w}}+\mathrm{M}_{\mathrm{B}} \mathrm{C}_{\mathrm{B}} \Delta \mathrm{T}_{\mathrm{B}}$
$=\frac{1}{2} \mathrm{M}_{\mathrm{B}} \mathrm{V}^{2}=\mathrm{M}_{\mathrm{w}} \mathrm{C}_{\mathrm{w}}\left(\Delta \mathrm{T}_{\mathrm{w}}\right)+\mathrm{M}_{\mathrm{B}} \mathrm{C}_{\mathrm{B}} \Delta \mathrm{T}_{\mathrm{B}}$
$\Rightarrow \frac{1}{2} \times 20 \times 10^{-3} \times 4 \times 10^{6}$
$\quad=(\Delta \mathrm{T})\left\{1 \times 3000+20 \times 10^{-3} \times 5000\right\}$
$\Rightarrow 40 \times 10^{3}=\Delta \mathrm{T}\{3000+100)$
$\Delta \mathrm{T}=\frac{40 \times 10^{3}}{3100}$
$\Delta \mathrm{T}=12.9$
$\mathrm{~T}_{\mathrm{f}}-25=12.9$
$\mathrm{~T}_{\mathrm{f}}=25+12.9=37.9^{\circ} \mathrm{C}$
$\mathrm{C}_{\mathrm{B}}=5000 \mathrm{~J} / \mathrm{Kg}-{ }^{\circ} \mathrm{C}$
$\mathrm{V}=2000 \mathrm{M} / \mathrm{s}$
$\mathrm{M}_{\mathrm{w}}=1 \mathrm{Kg}$
$\mathrm{C}_{\mathrm{w}}=3000 \mathrm{~J} / \mathrm{Kg}-{ }^{\circ} \mathrm{C}$
$\mathrm{T}_{\mathrm{f}}=25^{\circ} \mathrm{C}=298 \mathrm{~K}$
$\frac{1}{2} \mathrm{MV}^{2}=\mathrm{M}_{\mathrm{w}} \mathrm{C}_{\mathrm{w}} \Delta \mathrm{T}_{\mathrm{w}}+\mathrm{M}_{\mathrm{B}} \mathrm{C}_{\mathrm{B}} \Delta \mathrm{T}_{\mathrm{B}}$
$=\frac{1}{2} \mathrm{M}_{\mathrm{B}} \mathrm{V}^{2}=\mathrm{M}_{\mathrm{w}} \mathrm{C}_{\mathrm{w}}\left(\Delta \mathrm{T}_{\mathrm{w}}\right)+\mathrm{M}_{\mathrm{B}} \mathrm{C}_{\mathrm{B}} \Delta \mathrm{T}_{\mathrm{B}}$
$\Rightarrow \frac{1}{2} \times 20 \times 10^{-3} \times 4 \times 10^{6}$
$\quad=(\Delta \mathrm{T})\left\{1 \times 3000+20 \times 10^{-3} \times 5000\right\}$
$\Rightarrow 40 \times 10^{3}=\Delta \mathrm{T}\{3000+100)$
$\Delta \mathrm{T}=\frac{40 \times 10^{3}}{3100}$
$\Delta \mathrm{T}=12.9$
$\mathrm{~T}_{\mathrm{f}}-25=12.9$
$\mathrm{~T}_{\mathrm{f}}=25+12.9=37.9^{\circ} \mathrm{C}$
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