91Ó°ÊÓ

A physical quantity that expresses the hotness of matter is called temperature. It is one of the seven fundamental quantities. The SI unit for temperature is Kelvin \({\rm{(K)}}\).

The take-off speed of an airplane is \({\rm{250 km/h}}\). The brakes are applied, and the airplane comes to stop traveling a distance of \({\rm{2}}{\rm{.0 km}}\). \({\rm{85\% }}\) of the energy is converted into heat. Calculate the rise in temperature of the brakes if the specific heat of the brakes is \({\rm{800 J/kg}} \cdot {\rm{^\circ C}}\) and the mass of the airplane is \({\rm{1}}{{\rm{0}}^{\rm{5}}}{\rm{ kg}}\).

The loss of kinetic energy of the airplane is,

\(\begin{align}\Delta E &= \frac{1}{2}mv_i^2 - \frac{1}{2}mv_f^2\\ &= \frac{1}{2}m\left( {v_i^2 - v_f^2} \right)\end{align}\)

Here, \(m\)is the mass of the airplane, \({v_i}\) is the initial speed of the airplane, and \({v_f}\) is the final speed of the airplane.

Substitute \({\rm{1}}{{\rm{0}}^{\rm{5}}}{\rm{ kg}}\) for \(m\), \({\rm{250 km/h}}\) for \({v_i}\), and \({\rm{0}}\) for \({v_f}\),

\(\begin{align}E &= \frac{1}{2} \times \left( {{{10}^5} {\rm{kg}}} \right) \times \left( {{{\left( {250 {\rm{km}}/{\rm{h}}} \right)}^2} - {{\left( 0 \right)}^2}} \right)\\ &= \frac{1}{2} \times \left( {{{10}^5} {\rm{kg}}} \right) \times \left\{ {{{\left( {\left( {250 {\rm{km}}/{\rm{h}}} \right) \times \left( {\frac{{{{10}^3} {\rm{m}}}}{{1 {\rm{km}}}}} \right) \times \left( {\frac{{1 {\rm{h}}}}{{3600 {\rm{s}}}}} \right)} \right)}^2} - {{\left( 0 \right)}^2}} \right\}\\ &= 2.4 \times {10^8} {\rm{kg}} \cdot {{\rm{m}}^2}/{{\rm{s}}^2} \times \left( {\frac{{1{\rm{ J}}}}{{1{\rm{ kg}} \cdot {{\rm{m}}^2}/{{\rm{s}}^2}}}} \right)\\ &= 2.4 \times {10^8}{\rm{ J}}\end{align}\)

The amount of energy converted into heat is,

\(Q = \Delta E \times 85\% \)

Substitute \(2.4 \times 1{0^8} J\) for \(\Delta E\),

\(\begin{align}Q &= \left( {2.4 \times {{10}^8} {\rm{J}}} \right) \times 85\% \\ &= \left( {2.4 \times {{10}^8} {\rm{J}}} \right) \times \frac{{85}}{{100}}\\ &= 2.04 \times {10^8} {\rm{J}}\end{align}\)

The heat energy is,

\(Q = mc\Delta T\)

Here, \(Q\)is the heat energy, \(m\)is the mass of the airplane, \(c\)is the specific heat of breaks, and \(\Delta T\) is the rise in temperature.

Rearranging the above equation in order to get an expression for the rise in temperature.

\(\Delta T = \frac{Q}{{mc}}\)

Substitute \(2.04 \times 1{0^8} J\) for \(Q\), \({\rm{1}}{{\rm{0}}^{\rm{5}}}{\rm{ kg}}\) for \(m\), and \({\rm{800 J/kg}} \cdot {\rm{^\circ C}}\) for \(c\),

\(\begin{align}\Delta T &= \frac{{2.04 \times {{10}^8}{\rm{ J}}}}{{\left( {{{10}^5}{\rm{ kg}}} \right) \times \left( {800{\rm{ J}}/{\rm{kg}} \cdot ^\circ {\rm{C}}} \right)}}\\ &= 2.55^\circ {\rm{C}}\end{align}\)

Hence, the rise in the temperature is \({\rm{2}}{\rm{.55^\circ C}}\).