Question

A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.

What is the final temperature of the lead bullet? (Assume the bullet retains all heat.)

The melting point of lead is 327°C. The specific heat of lead is 0.128 J/g⋅°C.

The heat of fusion of lead is:
a. 227°Cb. 260°Cc. 293°Cd. 327°C

Answer 1

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

`KE = (1)/(2) mv^2`

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

`KE = (1)/(2) mv^2`

`KE = (1)/(2) (5*10^(-3))(300)^2`

`KE =  225J`

On the other hand the required Energy to heat up t melting point is

`Q_1 = mC_p \Delta T`

`Q_2 = L_f m`

Where,

m = Mass

`C_p =`Specific Heat

`\Delta T =`Change at temperature

`L_f =` Latent heat of fussion

Heat required to heat up to melting point,

`Q = Q_1+Q_2`

`Q = mC_p \Delta T+L_f m`

`Q = 5*0.128*(327-20) + 5*24.7`

`Q = 310J`

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.