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.

Answer and Explanation:

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.


Learn more about this topic:

Loading...
Kinetic Energy: Examples & Definition

from General Studies Science: Help & Review

Chapter 4 / Lesson 14
62K

Related to this Question

Explore our homework questions and answers library