Question

a 905 - g meteor impacts the earth at a speed of 1623 m/s. if all of its energy is entirely converted to heat in the meteor, what will be the resulting temperature rise of the meteor, assuming it does not melt? assume the meteor is made of iron.

Answer 1

Answer: 2859.78 k

Explanation: By using the law of conservation of energy, the kinetic energy of the meteor equals the heat energy.

Kinetic energy = 1/2mv^2

Heat energy = mcΔθ

Where m = mass of meteor , v = velocity of meteor = 1623 m/s

c = specific heat capacity of meteor (iron) = 460.548 j/kg/k

Δθ = change in temperature of meteor = ?

From law ofconservation of energy, we have that

1/2mv^2 = mcΔθ

By cancelling "m" on both sides, we have that

v^2/2 = cΔθ

v^2 = 2cΔθ

(1623)^2 = 2× 460.548 × Δθ

2634129 = 921.096 × Δθ

Δθ = 2634129 / 921.096

Δθ = 2859.78 k

Answer 2

The temperature rise of the meteor is 2859.78 k.

Given that:

• A 905 - g meteor impacts the earth at a speed of 1623 m/s.

Calculation of temperature:

$Kinetic energy = \frac{1}{2} mv^2\\\\Heat energy = mc\Delta\theta$

Here

m = mass of meteor ,

v = velocity of meteor = 1623 m/s

c = specific heat capacity of meteor (iron) = 460.548 j/kg/k

Now

$\frac{1}{2} mv^2 = mc\Delta\theta\\\\v^2\div 2 = c\Delta\theta\\\\v^2 = 2c\Delta\theta\\\\(1623)^2 = 2\times 460.548 \times \Delta\theta\\\\2634129 = 921.096 \times \Delta\theta\\\\\Delta\theta= 2634129 \div 921.096$

= 2859.78 k

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