Question

An ingot of lead is dropped with an initial velocity of 14 ft/s. When it hits the ground, its temperature has been increased by 0.25 oF. From what height was the ingot dropped (in ft)? Neglect air resistance. [h = 2.79 ft]

Answer 1

Answer:

The ingot was dropped at a height of 2.81 feet.

Explanation:

The ingot is released from a height $h$ and accelerated by gravity, it hits the ground and stops abruptly by impact, which contributes to the temperature increase of the material. We can model this systems by means of the First Law of Thermodynamics, which is generalized form of Principle of Energy Conservation:

$U_{g,1}-U_{g,2} +K_{1}-K_{2} = \Delta U_{sys}$ (Eq. 1)

Where:

$U_{g,1}$, $U_{g,2}$ - Initial and final gravitational potential energies, measured in pound-feet per second.

$K_{1}$, $K_{2}$ - Initial and final translational kinetic energies, measured in pound-feet per second.

$\Delta U_{sys}$ - Change in internal energy of the lingot of lead, measured in pound-feet per second.

By applying definitions of gravitational potential, translational kinetic and internal energies, we expand the expression above, eliminate mass in both sides and clear the final temperature of the ingot:

$g\cdot (z_{1}-z_{2})+ \frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2}) = c_{p}\cdot (T_{2}-T_{1})$

$z_{1}-z_{2}=\frac{c_{p}}{g} \cdot (T_{2}-T_{1})-\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2})$

If we know that $c_{p} = 753\,\frac{lbm\cdot ft}{lbm\cdot ^{\circ}F}$, $g = 32.174\,\frac{ft}{s^{2}}$, $T_{2}-T_{1} = 0.25\,^{\circ}F$, $v_{1} = 14\,\frac{ft}{s}$ and $v_{2} = 0\,\frac{ft}{s}$, the height from which the lingot dropped is:

$z_{1}-z_{2} = \left(\frac{753\,\frac{lbm\cdot ft}{lbm\cdot ^{\circ}F} }{32.174\,\frac{ft}{s^{2}} }\right)\cdot (0.25\,^{\circ}F)-\left[\frac{1}{2\cdot \left(32.174\,\frac{ft}{s^{2}} \right)} \right] \cdot \left[\left(14\,\frac{ft}{s} \right)^{2}-\left(0\,\frac{ft}{s} \right)^{2}\right]$

$z_{1}-z_{2} = 2.81\,ft$

The ingot was dropped at a height of 2.81 feet.