A 3.00 x 10^-3 kg lead bullet enters a target with a speed of 2.00 x 10^2 m/s and comes to rest...

Question:

A {eq}3.00 \times 10^{-3} {/eq} kg lead bullet enters a target with a speed of {eq}2.00 \times 10^2 {/eq} m/s and comes to rest within the target. Calculate the rise in temperature of the bullet assuming that 80.0% of the heat produced is absorbed by the bullet. The specific heat of lead is {eq}1.30 \times 10^2 {/eq} J/kg{eq}\cdot {/eq}K.

Energy.

If an object has kinetic energy by virtue of its movement and crashes into an obstacle, its energy is mainly transformed into internal energy which manifests itself in an increase in temperature of the object and the obstacle.

Answer and Explanation:

{eq}\text{Known data:}\\ m = 3.00\times{10^{-3 }}\,kg\\ v = 2.00\times{10^{2 }}\,m/s\\ c_{Pb} = 1.30\times{10^{2 }}{\rm \dfrac{J}{kg\,^oC} }\\ \text{Unknowns:}\\ \Delta T = ? \\ {/eq}

80% of the kinetic energy of the bullet is transformed into internal energy.

{eq}\Delta U = 0.800\left( {\dfrac{1}{2}m\,{v^2}} \right) = {\rm 0.800\left( {\dfrac{1}{2}\left( {3.00\times{10^{-3}}\,kg} \right){{\left( {2.00\times{10^{2}} \,m/s} \right)}^2}} \right) = 48\,J }\\ {/eq}

The increase of the internal energy of the bullet is equivalent to a heating without phase change.

{eq}Q = \Delta U = {\rm 48\,J}\\ Q = m\,{c_{Pb}}\,\Delta T \\ \Delta T = \dfrac{Q}{{m\,{c_{Pb}}}} = {\rm \dfrac{{48\,J}}{{\left( {3.00 \times {{10}^{ - 3}}kg} \right)\left( {1.30 \times {{10}^2}\dfrac{J}{{kg\,K}}} \right)}} = 123\,K{\text{ or }}123\,^oC }\\ {/eq}

The increase in the temperature of the bullet is:

{eq}\color{blue}{\Delta T = {\rm 123\,K}} {/eq}


Learn more about this topic:

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What is Energy Conservation? - Definition, Process & Examples

from ICSE Environmental Science: Study Guide & Syllabus

Chapter 1 / Lesson 6
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