# Question #8e674

Jan 30, 2017

(I) Kinetic energy $K E = \frac{1}{2} m {v}^{2}$,
where $m$ is mass of object and $v$ its velocity.

Kinetic energy of Allegheny class locomotive is given by, where given velocity has been converted to SI units
$K E = \frac{1}{2} \left(5.368 \times {10}^{5}\right) {\left(72.0 \times {10}^{3} / 3600\right)}^{2}$
$\implies K E = 1.0736 \times {10}^{8} J$
(II) Converting to CGS units,
$K E = 1.0736 \times {10}^{8} / 4.184 c a l$
$\implies 2.568421 \times {10}^{7} c a l$ .....(1)
Since it is steam engine as such we need to consider heat lost by
(1) steam at ${100}^{\circ} C$ to become water at ${100}^{\circ} C$
(2) water for $\Delta T = {75.0}^{\circ} C$

Let $m$ be water needed to provide required heat.
Heat lost in step (1) ${Q}_{1} = m L$
where $L = 539 c a l \cdot {g}^{-} 1$ is latent heat of vaporization of water
$\implies {Q}_{1} = 539 m \text{ } c a l$

Heat lost in step (2) ${Q}_{2} = m s \Delta T$
where $s = 1 c a l \cdot {g}^{-} 1 \cdot {K}^{-} 1$ is specific heat of water.
$\implies {Q}_{2} = m \times 1 \times 75.0$
$\implies {Q}_{2} = 75.0 m$
Total heat lost $= {Q}_{1} + {Q}_{2} = 539 m + 75.0 m = 614 m \text{ } c a l$ .....(2)

Assuming that there is no loss of heat and all lost heat is converted in the kinetic energy, equating (1) and (2)
$614 m = 2.568421 \times {10}^{7}$
$\implies m = \frac{2.568421 \times {10}^{7}}{614}$
$\implies m = 4.183 \times {10}^{4} g$