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henrco

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I believe my attempt below is correct but I'm not 100% sure, any guidance welcome.

Suppose a meteor of volume 1000 km3, density 5000 kg m-3 and speed 30000 km hr-1 crashed into the ocean and 10% of the impact energy was converted directly into heat.

i) Estimate the volume of water that would be evaporated.

ii) Given that the area of the oceans is 3.9 × 1015 m2, estimate the resulting change in their depth.

(The specific latent heat of vaporization for water is 2257 kJ kg-1 and the specific heat capacity of water is 4.18 kJ kg-1K-1. Assume that the mean ocean temperature is 4°C and that seawater has a density of 1000 kg m-3. Indicate clearly any further assumptions you make.)

i) Kinetic Energy = 0.5 mv^2

ii) Latent Heat Q = m l (m is the mass and l is the latent heat of vaporization)

iii) Heat Capacity = m c ΔT (m is the mass, ΔT is change in temp and c is the specific heat capacity)

Work the mass of the meteor : M = ρV = (5000)(1000x10^3) = 5x10^9 kg

Kinetic energy of meteor: 0.5(5x10^9)(3x10^7)^2 = 2.25x10^24 Joules

10% of kinetic energy turns to heat : 10%(2.25x10^24) = 2.25x10^23 Joules

ΔT (Change in temp required from 4C to 100C degrees or 277.15K to 373.15) : 96K

Energy required to heat 1kg of salt water by 96K.

Qheat = Mwater c ΔT = 1kg x (4.18) 96 = 401.28 kJ

Energy required to vapourize the salt water = 2257kJ/kg

Total energy required to heat and vapourize 1 kg of water = 2257 + 401.28 = 2658.28kJ

Determine the mass of water that is evaporated : 2.25x10^23 / 2658.28x10^3 = 8.46 x 10^16 kg

Convert this mass into volume: V = M/ρ = (8.46 x 10^16) / 1000 = 8.46 x 10^13 m3

1. Homework Statement1. Homework Statement

Suppose a meteor of volume 1000 km3, density 5000 kg m-3 and speed 30000 km hr-1 crashed into the ocean and 10% of the impact energy was converted directly into heat.

i) Estimate the volume of water that would be evaporated.

ii) Given that the area of the oceans is 3.9 × 1015 m2, estimate the resulting change in their depth.

(The specific latent heat of vaporization for water is 2257 kJ kg-1 and the specific heat capacity of water is 4.18 kJ kg-1K-1. Assume that the mean ocean temperature is 4°C and that seawater has a density of 1000 kg m-3. Indicate clearly any further assumptions you make.)

## Homework Equations

i) Kinetic Energy = 0.5 mv^2

ii) Latent Heat Q = m l (m is the mass and l is the latent heat of vaporization)

iii) Heat Capacity = m c ΔT (m is the mass, ΔT is change in temp and c is the specific heat capacity)

## The Attempt at a Solution

Work the mass of the meteor : M = ρV = (5000)(1000x10^3) = 5x10^9 kg

Kinetic energy of meteor: 0.5(5x10^9)(3x10^7)^2 = 2.25x10^24 Joules

10% of kinetic energy turns to heat : 10%(2.25x10^24) = 2.25x10^23 Joules

ΔT (Change in temp required from 4C to 100C degrees or 277.15K to 373.15) : 96K

Energy required to heat 1kg of salt water by 96K.

Qheat = Mwater c ΔT = 1kg x (4.18) 96 = 401.28 kJ

Energy required to vapourize the salt water = 2257kJ/kg

Total energy required to heat and vapourize 1 kg of water = 2257 + 401.28 = 2658.28kJ

Determine the mass of water that is evaporated : 2.25x10^23 / 2658.28x10^3 = 8.46 x 10^16 kg

Convert this mass into volume: V = M/ρ = (8.46 x 10^16) / 1000 = 8.46 x 10^13 m3

**Answer 1)**Volume of sea water vaporized = 8.46 x 10^13 m3**Answer 2)**Change in Ocean depth: (8.46 x 10^13)/ (3.9 x 10^15) = .022 m or 2.2cm
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