A 10.0 -g iron bullet with a speed of $4.00 \times 10^{2} \mathrm{m} / \mathrm{s}\( and a temperature of \)20.0^{\circ} \mathrm{C}$ is stopped in a \(0.500-\mathrm{kg}\) block of wood, also at \(20.0^{\circ} \mathrm{C} .\) (a) At first all of the bullet's kinetic energy goes into the internal energy of the bullet. Calculate the temperature increase of the bullet. (b) After a short time the bullet and the block come to the same temperature \(T\). Calculate \(T\), assuming no heat is lost to the environment.

The kinetic energy of the bullet is given by: \(KE = \frac{1}{2} mv^2\) where \(m\) is the mass of the bullet and \(v\) is its velocity. When the bullet comes to a stop, its kinetic energy is converted into internal energy, which increases the temperature of the bullet. We can calculate the temperature increase (\(\Delta T_{bullet}\)) using the energy conservation principle and the specific heat capacity of iron (\(c_{iron}\)): \(\Delta KE = m_{bullet} c_{iron} \Delta T_{bullet}\) The specific heat capacity of iron is \(c_{iron} = 450\ \text{J/kgK}\). We can rearrange the equation to solve for \(\Delta T_{bullet}\): \(\Delta T_{bullet} = \frac{\Delta KE}{m_{bullet} c_{iron}}\) Now plug in the given values and calculate the temperature increase.

When the bullet and the block of wood reach thermal equilibrium, their total energy remains constant. We can use the conservation of energy principle to relate the final temperatures and specific heat capacities of the bullet and the block: \(m_{bullet} c_{iron}(T-T_{bullet, initial}) = m_{wood} c_{wood}(T-T_{wood, initial})\) where \(m_{wood}\) is the mass of the block of wood and \(c_{wood}\) is its specific heat capacity. Since we assume that no heat is lost to the environment, the initial temperature for both the bullet and the block is \(20^{\circ}\mathrm{ C}\). We are given that \(m_{wood} = 0.500\ \text{kg}\), and the specific heat capacity of wood is typically around \(c_{wood} = 1700\ \text{J/kgK}\). Now, we can plug in the values and solve for the final equilibrium temperature \(T\). Note that we need to add the temperature increase calculated in Step 1 to the initial temperature of the bullet to find the current temperature \(T_{bullet, initial}\). After solving for the unknown variable \(T\), we will obtain the final equilibrium temperature.