A 6.5-g iron meteor hits the earth at a speed of 295 m/s. If its kinetic energy is entirely converted to heat in the meteor, by how much will its temperature rise? The specific heat of iron is $113 \mathrm { cal } / \mathrm { kg } \text {. }$ $\mathrm { C } ^ { \circ }$ and 1 cal=4.186 J

Question

A 6.5-g iron meteor hits the earth at a speed of  295 m/s. If its kinetic energy is entirely converted to heat in the meteor, by how much will its temperature rise? The specific heat of iron is $113 \mathrm { cal } / \mathrm { kg } \text {. }$ $\mathrm { C } ^ { \circ }$ and  1 cal=4.186 J

Quizplus · Accepted Answer

First, convert the mass of the meteor to kilograms: $6.5 \, \text{g} = 0.0065 \, \text{kg}$. Then, calculate the kinetic energy (KE) using the formula $KE = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the velocity. Substituting the given values, $KE = \frac{1}{2} \times 0.0065 \, \text{kg} \times (295 \, \text{m/s})^2 = 281.9625 \, \text{J}$.Convert the kinetic energy from joules to calories using the conversion factor $1 \, \text{cal} = 4.186 \, \text{J}$: $281.9625 \, \text{J} \times \frac{1 \, \text{cal}}{4.186 \, \text{J}} \approx 67.4 \, \text{cal}$.Finally, calculate the temperature rise ($\Delta T$) using the formula $\Delta T = \frac{Q}{mc}$, where $Q$ is the heat energy in calories, $m$ is the mass in kilograms, and $c$ is the specific heat capacity. Substituting the given values, $\Delta T = \frac{67.4 \, \text{cal}}{0.0065 \, \text{kg} \times 113 \, \text{cal/kg}^\circ C} \approx 92.0^\circ C$.

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