A 0.0500-kg lead bullet of volume 5.00 × 10 –6 m 3 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. Use the following information for lead:
-What additional heat would be needed to melt the bullet?
A)420 J
B)628 J
C)837 J
D)1160 J
E)2010 J

Question

A 0.0500-kg lead bullet of volume 5.00 × 10 –6 m 3 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.  Use the following information for lead: 
-What additional heat would be needed to melt the bullet?
A)420 J 
B)628 J 
C)837 J 
D)1160 J 
E)2010 J

Quizplus · Accepted Answer

The heat required to melt the bullet is given by the formula Q = mL, where m is the mass of the bullet and L is the latent heat of fusion of the bullet. The mass of the bullet is given as 10 g, and the latent heat of fusion of lead is 23.3 J/g. Therefore, the heat required to melt the bullet is Q = (10 g) x (23.3 J/g) = 233 J. The additional heat required to melt the bullet is therefore 233 J - 170 J = 63 J.

A 00500-Kg Lead Bullet of Volume 5