Question

A 0.0500-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?

Answer 1

Answer:

1964.8 J

$5.12894\times 10^{-6}\ m^3$

1150 Joules

Explanation:

m = Mass of bullet = 0.5 kg

$\Delta T$ = Change in temperature = (327-20)

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

$\beta$ = $84\times 10^{-6}\ /^{\circ}C$

$L_f$ = Latent heat of fusion of lead = $23000\ J/kg^{\circ}C$

(Values taken from properties of lead table)

Heat is given by

$Q=mc\Delta T\\\Rightarrow Q=0.05\times 128\times (327-20)\\\Rightarrow Q=1964.8\ J$

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

Change in volume is given by

$\Delta V=V_0\beta \Delta T\\\Rightarrow \Delta V=5\times 10^{-6}\times 84\times 10^{-6}\times (327-20)\\\Rightarrow \Delta V=1.2894\times 10^{-7}\ m^3$

$V=V_0+\Delta V\\\Rightarrow V=5\times 10^{-6}+1.2894\times 10^{-7}\\\Rightarrow V=5.12894\times 10^{-6}\ m^3$

The volume of the bullet when it comes to rest is $5.12894\times 10^{-6}\ m^3$

Heat needed for melting

$Q=mL_f\\\Rightarrow Q=0.05\times 23\times 10^3\\\Rightarrow Q=1150\ J$

The additional heat needed to melt the bullet is 1150 Joules