Solved Problem on Heat

A meteorite of mass 10 kg, when penetrating the Earth's atmosphere in a course of 3 km, has its speed reduced from 400 m/s to 200 m/s, due to air resistance. The trajectory can be considered a straight line. Considering the energy radiated by the meteorite to be negligible and its thermal conductivity perfect, determine:
a) The average force that acted on the meteorite;
b) The heat generated by friction during entry into the atmosphere;
c) Assuming all the heat produced is absorbed by the meteorite, and its initial temperature is 100 K, what is its final temperature?
Data:
Specific heat of the meteorite material: c = 0.1 kcal/kg.K and 1 kcal = 4200 J.

Problem data:

Meteorite mass: m = 10 kg;
Initial speed of the meteorite: v_i = 400 m/s;
Final speed of the meteorite: v_f = 200 m/s;
Path traveled in the atmosphere: ΔS = 3 km;
Specific heat of the constituent material of the meteorite: c = 0.1 kcal/kg.K;
Mechanical equivalent of heat: 1 kcal = 4200 J.

Problem diagram:

A meteorite of mass equal to 10 kg penetrates the atmosphere, and at a displacement of 3 km the velocity decreases from 400 m/s to 200 m/s, all the heat Q produced by friction with the atmosphere is absorbed by the meteorite itself (Figure 1).

Figure 1

Solution

First, we convert the space covered by the given meteorite in kilometers (km) to meters (m) used in the International System of Units (SI)

\[ \begin{gather} \Delta S=3\;\cancel{\mathrm{km}}\times\frac{1000\;\mathrm{m}}{1\;\cancel{\mathrm{km}}}=3000\;\mathrm{m} \end{gather} \]

a) The average force is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {\langle F\rangle =ma} \tag{I} \end{gather} \]

To determine the acceleration, we use the equation of velocity as a function of displacement

\[ \begin{gather} \bbox[#99CCFF,10px] {v^{2}=v_{0}^{2}+2a\Delta S} \end{gather} \]

using v₀ = v_i = 400 m/s, v = v_f = 200 m/s and ΔS = 3000 m

\[ \begin{gather} 200^{2}=400^{2}+2a\times 3000\\[5pt] 40000=160000+6000a\\[5pt] 6000a=40000-160000\\[5pt] a=-{\frac{120000}{6000}}\\[5pt] a=-20\;\mathrm{m/s}^{2} \tag{II} \end{gather} \]

substituting the mass given in the problem and the acceleration found above into equation (I)

\[ \begin{gather} \langle F\rangle =10\times(-20) \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\langle F\rangle =-200\mathrm{N}} \end{gather} \]

Note: The minus sign in acceleration indicates that the meteorite is decelerating, and in force indicates that the meteorite is under the action of the drag force due to friction with the atmosphere.

b) The change in energy, ΔE, between the initial kinetic energy, K_i, and the final kinetic energy, K_f, equals the heat produced by friction.
The kinetic energy is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {K=\frac{mv^{2}}{2}} \end{gather} \]

\[ \begin{gather} \Delta E=\left|K_{f}-K_{ci}\right|\\[5pt] \Delta E=\left|\frac{mv_{f}^{2}}{2}-\frac{mv_{i}^{2}}{2}\right|\\[5pt] \Delta E=\frac{m}{2}\left|v_{f}^{2}-v_{i}^{2}\right|\\[5pt] \Delta E=\frac{10}{2}\times\left|200^{2}-400^{2}\right|\\[5pt] \Delta E=5\times\left|40000-160000\right|\\[5pt] \Delta E=5\times120000\\[5pt] \Delta E=600000\;\mathrm{J} \end{gather} \]

Converting change in energy from joules (J) to kilocalories (kcal)

\[ \begin{gather} Q=\Delta E=600000\;\cancel{\mathrm{J}}\times\frac{1\;\mathrm{kcal}}{4200\;\cancel{\mathrm{J}}}\approx143\;\mathrm{kcal} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {Q\approx 143\;\mathrm{kcal}} \end{gather} \]

c) Assuming that the heat produced by the meteorite, calculated in item (b), is absorbed by it, we will have an increase in temperature.
The heat transfer equation is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {Q=mc\left(t_{f}-t_{i}\right)} \end{gather} \]

\[ \begin{gather} 143=10\times 0.1\times\left(t_{f}-100\right)\\[5pt] 143=1\times\left(t_{f}-100\right)\\[5pt] 143=t_{f}-100\\[5pt] t_{f}=143+100 \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {t_{f}=243\;\mathrm{K}} \end{gather} \]

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