Question

A lead bullet weighing 18.0g and travelling at 500 m/s is embedded in a wooden block of 1.00kg. If both the bullet and the block were initially at 25.0 ^oC, what is the final temperature of block containing bullet? Assume no temperature loss to the surroundings.

Answer 1

Solution : –

We have, kinetic energy of bullet is converted into heat  = 1/2 mv²

=  1/2 ×18×10^–3×(500)²

=  2.25×10³ J

=  (2.25×10³)/(4.184×10³) kcal

=  0.538 kcal

Also, mS = mS for bullet + mS for wood

= 18×10^–3×0.030 + 1×0.50

And  q = K.E.= mS∆T

Therefore,  ∆T  =  K.E./(mS)

=  0.538/(18×10^–3×0.030 + 1×0.50)

=  1.08 K

=  1.08° C

Therefore, final temperature  = 25.0° C + 10.8° C

=  26.08° C