Physics problem, I need help

A 10.2-g iron bullet with a speed of 300 m/s and a temperature of 20.0° C is stopped in a 0.500-kg block of wood, also at 20.0° 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.
° C

(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.
° C

(a)
Kinetic Energy = Heat Transfer (Q)
1/2 * m * v^2 = m * C * (Tf - Ti),

Solving for Tf

[v^2 / (2 * C)] + Ti = Tf

Tf = ?

(b)
Heat Transfer (Q block) = Heat Transfer (Q bullet) | where Tf will be the same for both and the Tf from part (a) will be used as Ti2 for the bullet.

M * C1 * (Tf - Ti1) = m * C2 * (Tf - Ti2)

Tf = [(M * C1 * Ti1) - (m * C2 * Ti2)] / (M * C1 - m * C2)

Your Welcome...

KE of bullet = ½mv²
= ½(10.2 E-3kg)(300)² .. ..►KE = 459 J

Assuming KE→Heat
459 J = m*S(iron)*∆θ
459 J = (10.2 E-3kg)(450 J/[kg.K]) ∆θ
►∆θ = 100ºC

Assuming a value S(wood) = 2000 J/(kg.K) (oak hardwood)
459J of KE = heat energy in bullet and wood ..
459 = (10.2E-3kg)(450)(∆θ) + (0.5kg)(2000)(∆θ)
459 = ∆θ (1004.6)
►∆θ = 0.46ºC

0.0051*300^2 is the bullet's kinetic energy in Joules

iron and wood specific heat capacity given by the link