A lead bullet weighing `18.0g` and travelling at `500 m//s` is embedded in a wooden block of `1.00kg`. If both the nullet and the block were initially at `25.0^(@)C`, what is the final temperature of the block containing bullet? Assume no temperature loss to the surrounding. (Heat capacity of wood `= 0.5 kcal kg^(-1) K^(-1)`, heat capacity of lead `= 0.030 kcal kg^(-1) K^(-1))`

Explanation

Kinetic enegry of bullet is converted into heat.
`KE = (1)/(2) mu^(2) = (1)/(2) xx 18 xx 10^(-3) xx (500)^(2)`
`= 2.25 xx 10^(3)J`
`= (2.25 xx 10^(3))/(4.184 xx 10^(3)) kcal = 0.538 kcal`
Also, `q = KE =mC DeltaT (C =` heat capacity) `:. DeltaT = (KE)/(mS)`
`mC = mC` for bullet `+mC` for wooden block
`= (18 xx 10^(-3) xx 0.030 +1 xx 0.50)`
`DeltaT = (0.538)/((18 xx 10^(-3)xx0.030 +1 xx 0.500)) = 1.08 K = 1.08^(@)C`
`:.` Final temperature `= (25.0 + 1.08) = 26.08^(@)C`