An iron meteorite melts when it enters the Earth’s atmosphere....

Question

An iron meteorite melts when it enters the Earth’s atmosphere. If its initial temperature was -105° C outside of Earth’s atmosphere, calculate the minimum velocity the meteorite must have had before it entered Earth’s atmosphere.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with heat. So in this problem, we have a copper fragment. It heats up and melts as it enters the atmosphere of Venus. Question wants to know what the minimum velocity must the fragment have had before entering the Venetian atmosphere. If its initial temperature was negative 110 °C, assuming the melting point of copper is 1100 °C. The specific heat capacity of copper is 390 joules per kilogram degrees Celsius. And the latent heat of fusion or copper is 2.1 multiplied by 10 to the five joules per kilogram given four possible choices as our answers. Choice A is 1.2 multiplied by 10 to the 3 m per second. Choice B is 2.4 multiplied by 10 to the 3 m per second. Point C is 5.4 multiplied by 10 to the 3 m per second. And choice D is 6.7 multiplied by 10 to 3 m per second. Now, we're told this problem that the copper fragment eats up and melts as it enters the atmosphere and the question wants to know what is the minimum velocity that has to have in order for this to happen. So you're gonna be dealing with heat, heat is a form of energy. So we also need to relate that to a energy that is related to motion. In this case, we're gonna be relating our heat to our kinetic energy. So that's gonna be our starting point. We'll have our connect and GK equal to the amount of heat that was transferred to the copper fragment. The first thing we need to do is recall our definition for kinetic energy that is one half MV squared. And that's gonna be equal to the total heat that's been transferred to copper fragment. For that, we need to recall two relationships. The first one is the heat that's transferred to the object in order to raise its temperature for a given mass though one half Q is equal to em. Delta P. So here C is our specific heat capacity M is the mass of the object and delta T is our change in temperature. That's gonna be one, that's gonna be one form of heat that we have. So that's let's start there. That's gonna have um For our first cue will have the specific heat capacity of copper multiplied by its mass multiplied by its change in temperature. So that'll be the final temperature minus the initial temperature. Now, we're gonna have to add to that the heat that's required to melt the object. So this first term in the heat um uh equation there gets us from negative 110 °C up to 1100 °C. So it heats the object up. But now we have to melt it. And for that, we need to remember the amount of heat that has to be transferred in order to do a phase change. So in this case, our Q is going to be equal to the massive object multiplied by the latent heat of fusion. The reason why we're using the latent heat of fusion as opposed the latent heat of vaporization is because we're going from a solid to a liquid, that's our latent heat of fusion. So we're gonna have to add to the other heat term. This M multiplied by LF there's our mass multiplied by our late heat fusion. So at this point I can solve for the speed. So I have, for my equation here, I have the one half MV squared is equal to DM multiplying the quantity of TF minus T I plus MLF. So has solved that for V, we'll have V is equal to I have the where route. And before I do anything, I noticed that the mass term cancels out for every term. So that gets rid of the math. We weren't giving any information about the math and the problem. So that's a good thing. And what I'm left with is um I'll have a factor of two multiplying the quantity of the multiplying the quantity of ef minus I llf Now, I have um values for all those quantities. So I can just go ahead and plug those in. I have V as equal to where it of two multiplying quantity or our specific heat capacity. C we were given a value of 390 joules per kilogram three °C and that gets multiplied by our final temperature, which is 1100. Are you Celsius minus our initial temperature, which is the negative 110 °C. And we want to add to that product, the late heat effusion or copper. And so we'll have plus the 2.1 multiplied by 10 to the five joules per kilogram. Everything on the right hand side is now just number. So I can plug that into my calculator and keeping just two significant figures. I get 1.2 multiplied by 10 to the 3 m per second. That's the answer I'm looking for. And that corresponds to answer a. So just a quick little recap here, um We actually had to use conservation of energy because we set the um total connect energy equal to the amount of heat that's been transferred to the fragment. And when we were looking at the heat, we actually had to conclude two different um terms for the heat. We had to first heat up the object from the initial temperature to the final temperature, then we had to melt the object. So I hope that this has been helpful and we'll see you in the next video.

An iron meteorite melts when it enters the Earth’s atmosphere. If its initial temperature was -105° C outside of Earth’s atmosphere, calculate the minimum velocity the meteorite must have had before it entered Earth’s atmosphere.

Key Concepts

Thermal Energy and Melting Point

Kinetic Energy and Velocity

Atmospheric Entry and Drag Force

Watch next