Meteor impact - Heat vaporization and capacity

[henrco]

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

1. 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

 

[SteamKing]

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

1. 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

How many cubic meters are in 1 cubic kilometer?

Remember, 1 cubic kilometer has the same volume as a cube which measures 1000 m on each edge.

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

What are the units of the velocity of the meteor on impact? Do those units produce joules for kinetic energy?

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

A temperature difference of 1° C = a temperature difference of 1° K.

There is no need to convert to absolute temperature when calculating temperature differences.

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

You've got some serious errors in your unit calculations which need fixing.

 

[henrco]

Thanks for your reply and for pointing out my mistakes.

Here are my corrections. For the last section "Energy required to heat 1kg..." you said I have serious unit calculations.
But apart from the ones that were carried through from my earlier mistakes I couldn't spot new ones?

I am however not sure about my answer to part 2) of the question.

The Attempt at a Solution

Convert the volume of the meteor from km^3 to m^3
1000 km^3 = 1x10^12 m

Work the mass of the meteor : M = ρV = (5000)(1x10^12) = 5x10^15 kg
Convert the velocity of the meteor from km/hr to m/s.
30,000 km/hr = 8333m/s
Kinetic energy of meteor: 0.5(5x10^15)(8333)^2 = 1.7x10^23 Joules10% of kinetic energy turns to heat : 10%(1.7x10^23) = 1.7x10^22 Joules

ΔT is difference in temperate from 4 to 100 degrees which is 96 degrees
Energy required to heat 1kg of salt water by 96 degrees
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 : (1.7x10^22) / (2658.28x10^3) = 6.53 x 10^15 kg
Convert this mass into volume: V = M/ρ = (6.53 x 10^15) / 1000 = 6.53 x 10^12m3

Answer 1) Volume of sea water vaporized = 6.53 x 10^12 m3

Answer 2) Change in Ocean depth: (6.53 x 10^12)/ (3.9 x 10^15) = 1.67x10^-3 m

 

[SteamKing]

Corrections look good.

The change in ocean depth due to vaporization of the water due to meteor impact < 2 mm.

 

[henrco]

Thanks for you help. Greatly appreciated.