Question

Question #8e674

Answer 1

(I) Kinetic energy `KE=1/2mv^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
`KE=1/2(5.368xx10^5)(72.0xx10^3/3600)^2`
`=>KE=1.0736xx10^8J`
(II) Converting to CGS units,
`KE=1.0736xx10^8/4.184cal`
`=>2.568421xx10^7cal` .....(1)
Since it is steam engine as such we need to consider heat lost by
(1) steam at `100^@C` to become water at `100^@C`
(2) water for `DeltaT=75.0^@C`

Let `m` be water needed to provide required heat.
Heat lost in step (1) `Q_1=mL`
where `L=539calcdotg^-1` is latent heat of vaporization of water
`=>Q_1=539m" "cal`

Heat lost in step (2) `Q_2=msDeltaT`
where `s=1calcdotg^-1cdotK^-1` is specific heat of water.
`=>Q_2=mxx1xx75.0`
`=>Q_2=75.0m`
Total heat lost `=Q_1+Q_2=539m+75.0m=614m" "cal` .....(2)

Assuming that there is no loss of heat and all lost heat is converted in the kinetic energy, equating (1) and (2)
`614m=2.568421xx10^7`
`=>m=(2.568421xx10^7)/614`
`=>m=4.183xx10^4g`