Question

A 6.30-g lead bullet traveling at 550 m/s is stopped by a large tree. If half the kinetic energy of the bullet is transformed into internal energy and remains with the bullet while the other half is transmitted to the tree, what is the increase in temperature of the bullet?

C

Answer 1

The kinetic energy of the bullet is

KE = (1/2) m v^2

The amount of heat, Q, available is

?Q = (1/2) KE = (1/4) m v^2

If c is the specific heat capacity and T is change in temperature then

?T = ?Q/(m c)

?T = (1/m c) x (1/4) m v^2 = (1/4) v^2/c

Notice that the mass of the bullet is irrelevant and cancels out, which makes sense. A bullet twice as massive carries twice as much KE, but there is twice as much mass to heat up, so the temperature change is the same. Assemble the numerical values in right units (kg, m, s)

c = 128 J/

Answer 2

avg Tkinetic_3k | 2'

High Temperature Low Temperature Heat transfer

An important idea related to temperature is the fact that a collision between a molecule with high kinetic energy and one with low kinetic energy will transfer energy to the molecule of lower kinetic energy. Part of the idea of temperature is that for two collections of the same type of molecules that are in contact with each other, the collection with higher average kinetic energy will transfer energy to the collection with lower average kinetic energy. We would say that the collection with higher kinetic energy has a higher temperature, and that net energy transfer will be from the higher temperature collection to the lower temperature collection, and not vice versa. Clearly, temperature has to do with the kinetic energy of the molecules, and if the molecules act like independent point masses, then we could define temperature in terms of the average translational kinetic energy of the molecules, the so-called "kinetic temperature".

so KE = 302500 *3.15/1000 = 952875/1000=952.875

so T=2/(3k) *952.875