Question

A 0.02kg lead bullet traveling 200 m/s

strikes an armor plate 2kg at rest. Assume
all its kinetic enrgy transfer to heat what is
the change in temperature? the specific heat
of lead is for lead is 130 J/kg.C

Answer 1

`\Delta T=1.53^(\circ)`

Step-by-step explanation:

Given that,

Mass of a lead bullet, m = 0.02 kg

Speed of the bullet, v = 200 m/s

It strikes with an armor plate 2kg at rest. All its kinetic energy is transferred to heat. We need to find the change in temperature.

A/c to the law of conservation of energy,

`(1)/(2)mv^2=Mc\Delta T\\\\\Delta T=(mv^2)/(2Mc)\\\\\Delta T=(0.02* 200^2)/(2* 2* 130)\\\\\Delta T=1.53^(\circ)`

So, the required change in temperature is equal to
`1.53^(\circ)`.