Question

a lead bullet of mass 500 grams.(c=0.3 CGS units )moving at a speed of 210 m/s is suddenly stopped. Find the increase in temperature.

Answer 1

Let:

$$V = 210 \, \mathrm{m/s}$$ $$m = 500g = 0,5 \, \mathrm{Kg}$$ $$c = 0,3 \, \mathrm{CGS} \, \mathrm{units} = 300 \, \mathrm{J/KgK}$$ $$\Delta T = ?$$

The bullet kinetic energy passes in heat:

$$Q = E k$$ $$C m \Delta t = \frac{m V^2}{2}$$ $$\Delta t = \frac{m V^2}{2 C m} = \frac{V^2}{2 C}$$

Let's enter the data:

$$\Delta t = \frac{210^2}{2 \times 300} = 73,5{}^\circ$$

Answer:

The increase in temperature is: $73,5{}^\circ$