100 grams of ice, with a temperature of -10°C, is added to a styrofoam cup of water. The water is initially at +10°C, and has an unknown mass m. If the final temperature of the mixture is 0°C, what is the unknown mass m? Assume that no heat is exchanged with the cup or with the surroundings.

Use these approximate values to determine your answer:

Specific heat of liquid water is about 4000 J/(kg °C)
Specific heat of ice is about 2000 J/(kg °C)
Latent heat of fusion of water is about 3 x 10⁵ J/kg

The first thing to realize is that there is not enough information given in the problem to determine a single value of m, because we don't know how much ice and how much water is in the cup at equilibrium.

One possible starting point is to determine what happens if nothing changes phase. How much water at +10°C does it take to bring 100 g of ice at -10°C to 0°C? (The water also ends up at 0°C.)

You can do heat lost = heat gained or the equivalent method:

ΣQ = 0

m_ice c_ice ΔT_ice + m_water c_water ΔT_water = 0

Plugging in numbers gives:

(100 g)*[2000 J/(kg °C)]*(10°C) + m*[4000 J/(kg °C)]*(-10°C) = 0

Lot's of things cancel and we're left with:

100 g = 2m, so m = 50 g.

So, that's one possible answer.

How would we figure out the maximum possible value of m? What is the largest mass of water at +10°C we could add to 100 g of ice at -10°C and still have an equilibrium temperature of 0°C?

In this case all the ice would melt, so let's set up the equation that reflects that:

m_ice c_ice ΔT_ice + m_ice L_f + m_water c_water ΔT_water = 0

(100 g)*[2000 J/(kg °C)]*(10°C) + (100 g)*(3 x 10⁵ J/kg) + m_max*[4000 J/(kg °C)]*(-10°C) = 0

Multiplying things out and dividing by a factor of 10000 gives:

200 g + 3000 g = 4m_max

m_max = 3200/4 = 800 g.

So, we have determined a couple of different values of the unknown mass m. Are we done? In other words, is the answer that as long as m is greater than or equal to 50 g or less than or equal to 800 g the final temperature of the mixture will be 0°C, and for all values of m outside of this range the final temperature will be something other than 0°C?

1. Yes, that is correct
2. No, at some values of m between 50 and 800 g the final temperature will be something other than 0°C.
3. No, some values below 50 g would also work.
4. No, some values above 800 g would also work.

What would happen if the original mass of water was larger than 800 g?

All the ice would melt and the final temperature would be more than 0°C.

But, if the mass of water was less than 50 g we could still end up with a temperature of 0°C because some (or all) of the water could freeze. What is the smallest value of m we could use and still have a final temperature of 0°C?

In this case all the water would freeze, so let's set up the equation that reflects that:

m_ice c_ice ΔT_ice + m_water c_water ΔT_water - m_water L_f = 0

(100 g)*[2000 J/(kg °C)]*(10°C) + m_min*[4000 J/(kg °C)]*(-10°C) - m_min*(3 x 10⁵ J/kg) = 0

Multiplying things out and dividing by a factor of 10000 gives:

200 g = 4m_min + 30m_min

m_min = 200/34, which is about 6 g.

So, amazingly enough, we can use anywhere from about 6 g of water to 800 g of water and still end up with a final temperature of 0°C.