# A lead bullet strikes a target with a velocity of 200 m / s and is brought to rest. Calculate the...

## Question:

A lead bullet strikes a target with a velocity of {eq}200 \ m / s {/eq} and is brought to rest. Calculate the rise in temperature of the bullet assuming that all heat developed remains in the bullet. Specific heat of lead is {eq}127.7 \ J / kg \cdot K. {/eq}

## Heat and Temperature Change

When a quantity {eq}Q {/eq} of heat is supplied to an object of mass {eq}m {/eq}, made of a substance with specific heat capacity {eq}c {/eq}, it will manifest a temperature rise expressed as:

{eq}\displaystyle \Delta T = \frac{Q}{mc} {/eq}.

An analogous process is true for heat removed from an object, where the temperature drops.

All the kinetic energy possessed by the lead bullet is converted to heat, which is reported to remain within the bullet itself. Let's first quantify this kinetic energy:

{eq}\displaystyle \begin{align} K &= \frac{1}{2}(m)v^2\\ &= \frac{1}{2}(m) (200)^2 \\ &= 20000 m \ J \end{align} {/eq}.

Above, the mass of the bullet is denoted by {eq}m {/eq} and its velocity by {eq}v {/eq}.

Now, given the specific heat capacity of lead as {eq}c = 127.7 \ J/kgK {/eq},

we can directly compute the rise in temperature in the bullet as:

{eq}\displaystyle \begin{align} \Delta T &= \frac{K}{mc} \\ &= \frac{20000 \ m}{m \times 127.7} \\ &= 156.6 \ K \end{align} {/eq}.

Note that in this question, we did not need to know the mass of the bullet explicitly. 