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

The calculation involves determining the heat required to increase the bullet's temperature, calculating the change in volume due to thermal expansion, and the additional heat required to melt the bullet, using the specific heat capacity, coefficient of thermal expansion, and latent heat of fusion for lead.

Step-by-step explanation:

To solve this problem, we will use the specific heat capacity formula and the properties of lead to calculate the heat needed for temperature change and melting.

Heat required to raise the bullet's temperature: The heat (Q) needed can be calculated using the specific heat capacity equation Q = mcΔT, where 'm' is mass, 'c' is specific heat capacity, and ΔT is the change in temperature. For lead, the specific heat capacity (c) is approximately 128 J/(kg·°C).

Volume: To find the volume at the final temperature, we use the formula V = V_0(1+ αΔT), where V_0 is the initial volume, α is the coefficient of thermal expansion for lead, and ΔT is the change in temperature.

Additional heat to melt the bullet: To calculate the additional heat (Q_m) required to melt the bullet, we use the formula Q_m = mL_f, where 'm' is the mass of the bullet and L_f is the latent heat of fusion for lead.

To calculate these values, we also need the initial temperature (T_i), final temperature (T_f), and coefficient of thermal expansion for lead, which is 29.3 × 10⁻⁶ °C-1. For melting lead, the latent heat of fusion (L_f) is 24.7 kJ/kg.

Answer 2

1964.8 J

`5.12894* 10^(-6)\ m^3`

1150 Joules

Step-by-step 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* 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* 128* (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* 10^(-6)* 84* 10^(-6)* (327-20)\\\Rightarrow \Delta V=1.2894* 10^(-7)\ m^3`

`V=V_0+\Delta V\\\Rightarrow V=5* 10^(-6)+1.2894* 10^(-7)\\\Rightarrow V=5.12894* 10^(-6)\ m^3`

The volume of the bullet when it comes to rest is
`5.12894* 10^(-6)\ m^3`

Heat needed for melting

`Q=mL_f\\\Rightarrow Q=0.05* 23* 10^3\\\Rightarrow Q=1150\ J`

The additional heat needed to melt the bullet is 1150 Joules