A0.0500-kg lead bullet of volume 5.00 × 10–6 m3 at 20.0 °c hits a block that is made of an ideal thermal insulator and comes to rest at its center. at that time, the temperature of the bullet is 327 °c.

a. how much heat was needed to raise the bullet to its final temperature? b. what is the volume of the bullet when it comes to rest?
c. what additional heat would be needed to melt the bullet?

1964.8 J

1150 Joules

Explanation:

m = Mass of bullet = 0.5 kg

= Change in temperature = (327-20)

c = Specific heat of lead = 128 J/kg °C

=

= Latent heat of fusion of lead =

(Values taken from properties of lead table)

Heat is given by

The heat needed to raise the bullet to its final temperature is 1964.8 J

Change in volume is given by

The volume of the bullet when it comes to rest is

Heat needed for melting

The additional heat needed to melt the bullet is 1150 Joules

it would be kevlar ^^