Question

A bullet of mass 5g, travelling with a speed of 210m/s, strikes a fixed wooden target. One half of its kinetics energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is 0.030cal/(g- °C) (1cal = 4.2 × 107 ergs) close to:

• A. 87.5°C
• B. 83.3°C
• C. 119.2°C
• D. 38.4°C

Answer 1

Answer: (A)

87.5°C

Answer 2

Here's a breakdown of the solution:

1. Calculate the initial kinetic energy (KE) of the bullet using the formula KE = $\frac{1}{2}mv^2$, ensuring mass is in kg and velocity in m/s to get KE in Joules.

2. Determine the heat converted into thermal energy in the bullet, which is half of the initial KE.

3. Convert the specific heat from cal/(g-°C) to J/(g-°C) using the given conversion factor (1 cal = 4.2 J).

4. Use the heat formula $Q = mc\Delta T$, where Q is the heat absorbed, m is the mass in grams, c is the specific heat in J/(g-°C), and $\Delta T$ is the temperature change.

5. Solve for $\Delta T$.