I don't think the question can be answered because you don't say how the orbital energy is to be dissipated. However it's quite interesting to compare the orbital energy with the energy required to boil the ice.

Let's suppose our ice supplied is aboard the International Space Station, so they are at an altitude of $h$ = 300km and moving at an orbital velocity of about $v_o$ = 7.7km/sec. At the latitude of New Mexico (34°N) the Earth's surface is moving at about $v_e$= 370m/sec. So the change in kinetic energy is:

$$\begin{align} \Delta T &= \tfrac{1}{2}m v_o^2 - \tfrac{1}{2}m v_e^2 \\ &= 29.6\space\text{Mj/kg} \end{align}$$

The change in potential energy is:

$$\begin{align} \Delta U &= \frac{GM}{r_e} - \frac{GM}{r_e + h} \\ &= 3.1\space\text{Mj/kg} \end{align}$$

So the total energy change in bringing 1kg of ice from the ISS to New Mexico is:

$$ \Delta E = \Delta T + \Delta U = 32.7 \text{MJ/kg} $$

Could we use this energy to boil off some of 1kg of ice and leave the rest available for cooling drinks? Well suppose we start with the ice at absolute zero (it's cold in space) and see how much energy it takes to boil it. The constants we need are:

$$\begin{align} \text{Specific heat of ice (-10C)} &= 2000 \space \text{J/kg.K} \\ \text{Latent heat of fusion} &= 334000 \space \text{J/kg} \\ \text{Specific heat of water} &= 4200 \space \text{J/kg.K} \\ \text{Latent heat of vap.} &= 2257000 \space \text{J/kg.K} \end{align}$$

Assuming these constants don't change with temperature$^1$ the energy required to turn 1kg of ice at absolute zero to a kg of steam at 100°C is:

$$\begin{align} \Delta E &= 2000*273 + 334000 + 4200*100 * 2257000 = 0 \\ &= 3.56 \space \text{MJ/kg} \end{align}$$

So the energy required to bring 1 kg of ice to rest in New Mexico is about ten times the amount of energy needed to boil away the ice even starting from absolute zero. You're going to have to find some other way of dissipating the energy.

$^1$ the specific heat of ice decreases with falling temperature so the energy calculated to boil the ice is a slight overestimate.

I don't think the question can be answered because you don't say how the orbital energy is to be dissipated. However it's quite interesting to compare the orbital energy with the energy required to boil the ice.

Let's suppose our ice supplied is aboard the International Space Station, so they are at an altitude of $h = 300\ \mathrm{km}$ and moving at an orbital velocity of about $v_\mathrm o = 7.7\ \mathrm{km/s}$. At the latitude of New Mexico (34° N) the Earth's surface is moving at about $v_\mathrm e = 370\ \mathrm{m/s}$. So the change in kinetic energy is:

$$\begin{align} \Delta T &= \tfrac{1}{2}m v_\mathrm o^2 - \tfrac{1}{2}m v_\mathrm e^2 \\ &= 29.6\ \mathrm{MJ/kg} \end{align}$$

The change in potential energy is:

$$\begin{align} \Delta U &= \frac{GM}{r_\mathrm e} - \frac{GM}{r_\mathrm e + h} \\ &= 3.1\ \mathrm{MJ/kg} \end{align}$$

So the total energy change in bringing $1\ \mathrm{kg}$ of ice from the ISS to New Mexico is:

$$ \Delta E = \Delta T + \Delta U = 32.7\ \mathrm{MJ/kg} $$

Could we use this energy to boil off some of $1\ \mathrm{kg}$ of ice and leave the rest available for cooling drinks? Well suppose we start with the ice at absolute zero (it's cold in space) and see how much energy it takes to boil it. The constants we need are:

$$\begin{align} \text{Specific heat of ice}\ (-10\ \mathrm{^\circ C}) &= 2\,000\ \mathrm{J/(kg\ K)} \\ \text{Latent heat of fusion} &= 334\,000\ \mathrm{J/kg} \\ \text{Specific heat of water} &= 4\,200\ \mathrm{J/(kg\ K)} \\ \text{Latent heat of vap.} &= 2\,257\,000\ \mathrm{J/kg} \end{align}$$

Assuming these constants don't change with temperature$^1$ the energy required to turn $1\ \mathrm{kg}$ of ice at absolute zero to a $\mathrm{kg}$ of steam at $100\ \mathrm{^\circ C}$ is:

$$\begin{align} \Delta E &= 2\,000\ \mathrm{J/(kg\ K)}\times273\ \mathrm K + 334\,000\ \mathrm{J/kg} + 4\,200\ \mathrm{J/(kg\ K)}\times100\ \mathrm K + 2\,257\,000\ \mathrm{J/kg} \\ &= 3.56\ \mathrm{MJ/kg} \end{align}$$

So the energy required to bring $1\ \mathrm{kg}$ of ice to rest in New Mexico is about ten times the amount of energy needed to boil away the ice even starting from absolute zero. You're going to have to find some other way of dissipating the energy.

$^1$ the specific heat of ice decreases with falling temperature so the energy calculated to boil the ice is a slight overestimate.

I don't think the question can be answered because you don't say how the orbital energy is to be dissipated. However it's quite interesting to compare the orbital energy with the energy required to boil the ice.

Let's suppose our ice supplied is aboard the International Space Station, so they are at an altitude of $h$ = 300km and moving at an orbital velocity of about $v_o$ = 7.7km/sec. At the latitude of New Mexico (34°N) the Earth's surface is moving at about $v_e$= 370m/sec. So the change in kinetic energy is:

$$\begin{align} \Delta T &= \tfrac{1}{2}m v_o^2 - \tfrac{1}{2}m v_e^2 \\ &= 29.6\space\text{Mj/kg} \end{align}$$

The change in potential energy is:

$$\begin{align} \Delta U &= \frac{GM}{r_e} - \frac{GM}{r_e + h} \\ &= 3.1\space\text{Mj/kg} \end{align}$$

So the total energy change in bringing 1kg of ice from the ISS to New Mexico is:

$$ \Delta E = \Delta T + \Delta U = 32.7 \text{MJ/kg} $$

Could we use this energy to boil off some of 1kg of ice and leave the rest available for cooling drinks? Well suppose we start with the ice at absolute zero (it's cold in space) and see how much energy it takes to boil it. The constants we need are:

$$\begin{align} \text{Specific heat of ice (-10C)} &= 2000 \space \text{J/kg.K} \\ \text{Latent heat of fusion} &= 334000 \space \text{J/kg} \\ \text{Specific heat of water} &= 4200 \space \text{J/kg.K} \\ \text{Latent heat of vap.} &= 2257000 \space \text{J/kg.K} \end{align}$$

Assuming these constants don't change with temperature the energy required to turn 1kg of ice at absolute zero to a kg of steam at 100°C is:

$$\begin{align} \Delta E &= 2000*273 + 334000 + 4200*100 * 2257000 = 0 \\ &= 3.56 \space \text{MJ/kg} \end{align}$$

So the energy required to bring 1 kg of ice to rest in New Mexico is about ten times the amount of energy needed to boil away the ice even starting from absolute zero. You're going to have to find some other way of dissipating the energy.

I don't think the question can be answered because you don't say how the orbital energy is to be dissipated. However it's quite interesting to compare the orbital energy with the energy required to boil the ice.

Let's suppose our ice supplied is aboard the International Space Station, so they are at an altitude of $h$ = 300km and moving at an orbital velocity of about $v_o$ = 7.7km/sec. At the latitude of New Mexico (34°N) the Earth's surface is moving at about $v_e$= 370m/sec. So the change in kinetic energy is:

$$\begin{align} \Delta T &= \tfrac{1}{2}m v_o^2 - \tfrac{1}{2}m v_e^2 \\ &= 29.6\space\text{Mj/kg} \end{align}$$

The change in potential energy is:

$$\begin{align} \Delta U &= \frac{GM}{r_e} - \frac{GM}{r_e + h} \\ &= 3.1\space\text{Mj/kg} \end{align}$$

So the total energy change in bringing 1kg of ice from the ISS to New Mexico is:

$$ \Delta E = \Delta T + \Delta U = 32.7 \text{MJ/kg} $$

Could we use this energy to boil off some of 1kg of ice and leave the rest available for cooling drinks? Well suppose we start with the ice at absolute zero (it's cold in space) and see how much energy it takes to boil it. The constants we need are:

$$\begin{align} \text{Specific heat of ice (-10C)} &= 2000 \space \text{J/kg.K} \\ \text{Latent heat of fusion} &= 334000 \space \text{J/kg} \\ \text{Specific heat of water} &= 4200 \space \text{J/kg.K} \\ \text{Latent heat of vap.} &= 2257000 \space \text{J/kg.K} \end{align}$$

Assuming these constants don't change with temperature$^1$ the energy required to turn 1kg of ice at absolute zero to a kg of steam at 100°C is:

$$\begin{align} \Delta E &= 2000*273 + 334000 + 4200*100 * 2257000 = 0 \\ &= 3.56 \space \text{MJ/kg} \end{align}$$

So the energy required to bring 1 kg of ice to rest in New Mexico is about ten times the amount of energy needed to boil away the ice even starting from absolute zero. You're going to have to find some other way of dissipating the energy.

$^1$ the specific heat of ice decreases with falling temperature so the energy calculated to boil the ice is a slight overestimate.