Engineering
Physics
Calorimetry
Question
A lead bullet strikes against a steel plate with a velocity 200 ms–1. If the impact is perfectly inelastic and the heat produced is equally shared between the bullet and the target, then the rise in temperature of the bullet is (specific heat capacity of head = 125 Jkg–1K–1)
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Solution
\[ \begin{aligned} \text{Initial KE of bullet} &= \frac{1}{2}mv^2 \\ &= \frac{1}{2}m(200)^2 \\ &= 20000m \\[6pt] \text{Heat gained by bullet} &= \frac{1}{2}\times KE \\ &= 10000m \\[6pt] Q &= mc\Delta T \\ 10000m &= m(125)\Delta T \\ \Delta T &= \frac{10000}{125} \\ \Delta T &= 80\ \text{C} \end{aligned} \]