# 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

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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))

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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