What is the temperature of the ice water mixture

Temperature of ice-water mixture

Homework Statement

A beaker containing 50g of ice and 250g of liquid water is initially at 0celsius. It is then heated on a hot plate until half of the ice is melted. What is the temperature of the ice-water mixture at that point?
A. 2.0 C
B. 0.5 C
C. 1.0 C
D. 1.5 C
E. 0.0 C

Homework Equations

The Attempt at a Solution

Answers and Replies

What is true about ice-water mixtures that can help you know their temperature?

I have another related question, so I’m going to post it here instead of creating a new topic:

b. The ice cube is added to 2.5 kg of water at 25° C. What is the final temperature of the equilibrium mixture?

c. If 720 g of ice water is added instead to the 2.5 kg of water at 25° C, what is the final equilibrium temperature of the mixture?

The specific heat of ice and water are not the same, so ice and water will need separate terms in the equation.

For part a there are three processes that occur: first the ice warms up to the freezing point, then it melts, then the water warms up to the final termperature.

If you don’t mind, please post the numbers that you use in your calculation (the way you did in post #7). It makes it much easier to see what you did.

Your statments about energy gained and energy lost sound good. But I’m not sure about the 1.5 degrees.

You have a small typo right after the 333000; it should be a plus sign, but you probably entered into your calculator correctly.

So when I check the number by calculating the left side with T=1.5, I get:

[2220 J/kgC * 0.720kg * 0-(-10)] + [0.720kg * 333000 J/kg + 4190J/kgC * 0.720kg * (T-0)]

=15984+239760 + 4525.2 = 260269

and the right side is

4190 J/kgC * 2.5kg * (25-T) = 4190 * 2.5 (25-1.5) = 246163

Since the right and left sides don’t equal, that’s why I thought 1.5 degrees was not correct, and I think you made an algebraic error when solving for T.

Combing the multiplications in your expression on the left side and multiplying out the terms on the right side gives:

15984+239760 +3016.8 T =261875-10475 T

which you can then solve for T. Is that what you are getting?

LPI | Education

Earth and Space Science News for Educators

That’s A N(ice) Temperature

Overview

In this 30-minute companion activity to The Melting Point, teams of children ages 8 to 13 experiment to find the melting and freezing points of water and ice.

What’s the Point?

Materials

For each child

For each group of four to six children:

For the facilitator:

Preparation

Activity

Invite the children to record their predictions in their Ice Investigator Journals.

After the thermometer has been in the ice for a few minutes, the children can attempt to get a temperature of their ice. The readings they collect may be close to 32°F (0°C), but may vary for several reasons. The warmer air in the room can circulate in the air pockets between the pieces of crushed ice, causing their readings to be warmer. Warmer readings may also arise because the thermometer was in contact with the sides or bottom of the cup, or it wasn’t buried deeply enough in the ice, or the children may have held it out of the ice for too long before reading it. Colder reading may occur because standard household freezers are generally kept at about 0°F (-18°C) and the ice will also be that cold. In addition, some thermometers read differently.

Facilitator’s Note: Adding salt lowers the melting point of water.

Pure water and ice, kept insulated from the warm outside world, come to equilibrium over time.В On a molecular level, water molecules are freezing onto the ice at the same rate as they are melting off of it. The entire water/ice solution is at the melting/freezing point, 32°F (0°C). Adding rock salt — or any substance that dissolves in water — disrupts this equilibrium. Fewer water molecules are interacting with the ice at any given moment, so the freezing rate is slowed. The salt has no effect on the melting rate, so more melting occurs than freezing — melting «wins» — and the ice melts. In doing so, heat energy is used to break the hydrogen bonds that hold the molecules in the ice together. In other words, the ice «uses up» some warmth from the solution and the temperature drops. Melting and freezing again match rates («tie») once the temperature has dropped to the new melting point.

The greater the amount of salt, the lower the freezing point (to a point; once there is sufficient salt that no more will dissolve, the freezing point no longer decreases). Ocean water is about 3.5% salt; sea water freezes at about 28°F (-2°C). A 10% salt solution freezes at about 20°F (-6°C), and a 20% solution freezes at 2°F (-16°C).

Conclusion

If the children have started to construct a snow mobile, invite them to record any answers they discovered on the appropriate pieces. Some of the questions that they can answer might include:

Have them annotate any new questions they have or interesting things they learned on the appropriate shapes of either raindrop, cloud, or snowflake.

What is the temperature of the flowing water in icy river?

Boundaries of the question:

The question itself:

What is the temperature of the flowing water in icy river?
More precisely, does the temperature of the water drop below 0 °C?
If it does, is this common? If it doesn’t, does it get close to it at all?

Attempted research to find the answer :

I had in mind that the current allowed the water to stay liquid, but that might be wrong because there is no proof to it. However, this answer says:

For flowing water to freeze, the temperature would have to be exceptionally cold.

Which seems to agree that current can lower the freezing point.

I found a great document about the temperature of the water within lake, but didn’t find any reliable answer for rivers.

EDIT
I have come to believe that temperatures don’t typically fall below 0 °C, based upon aggregating the following information.

Earth is warming the water from below.

Using this, we can probably conclude that water won’t cool down enough during it’s journey down the river. Thus, the water would need to be already colder at the start of the journey. We know that the start is a lake with calm water. So, it’s unlikely to happen because the lake would freeze completely and stop flowing out into the river.

1 Answer 1

This is a complex question.

I will assume the following to be able to frame an answer properly:

This all means that the cold region’s rivers are not thermally static through time. Even in the High Arctic, an icy river can be frozen along both its vertical and horizontal profile during a good part of the long winter, and thaw completely during the short summer. But other considerations exists.

Let’s examine what can happen when the snow thaw is sometime in June. The discharge signature for a river in the High Arctic is nival, with a discharge peak occurring during early summer due to the massive input of snowmelt water, and additional weaker responses from the small rain amounts. This looks like this:

Source: Woo (2012) Permafrost Hydrology, fig 10.14; from river McMaster, 74° N.

So what basically is happening here?

During the next several months, snow will accumulate in the landscape. In a landscape there are snow sources and snow sinks (Pomeroy et al. 1997); sources may be higher lands, outcrops, plateaus, or exposed areas, and sinks lower areas, such as depression or channels. Our typical nival river will be a sink in this context.

What is happening in June, in more detail?
Let’s analyse this further with a more detailed (though raw and unprocessed) graph from a small stream in a gully near Pond Inlet, with resolution available up to the nearest day:

Source: Stream temperature and water pressure in a stream located in a gully on Bylot Island, 85 km from Pond Inlet; raw, partial data from my PhD work. Temperature in blue, scale on the right, and pressure in black, scale on the left. Dates on the x-axis.

When a critical mass of water accumulates near the logger emplacement in the stream channel, the bulk of the snow melt and water then shifts to positive temperatures, still near zero, with diurnal/nocturnal shifts (blue line becoming wave-like).

Note: This small stream is minimally impacted by the geothermal flux (heat flowing up from the ground below).

Summary, in relation to the questions:

What is the temperature of the ice water mixture

Problem #1: Determine the final temperature when 10.0 g of steam at 100.0 °C mixes with 500.0 grams of water at 25.0 °C.

This problem is like 9 and 10 in Worksheet #2 with one difference. The sample problem has steam and the worksheet problem have ice. The ONLY differences are the numbers used and the fact that ice is associated with «cold water» side of the equation. The techniques are the same.

The «warm water» in this case is going to do two things:

a) as a gas, condense at 100.0 °C to liquid water
b) as a liquid, the temperature goes down some unknown amount

The colder water will receive all the energy for this to happen. Keeping in mind that the heat lost by the warmer must equal the heat gained by the colder, we have this:

heat lost to cooler water by condensing steam + heat used to warm cooler water = heat gained by the cooler water

Please note again that there are TWO sources of heat energy (steam condensing, then the warm water cooling down). The total energy lost by both will equal the heat energy gained by the cooler water.

Here are the numbers:

Notice I used 40700 J rather than 40.7 kJ for the molar heat of vaporization.

Proceeding with the solution, I get:

and so, x = 37.1 °C.

Keep in mind that ‘x’ was identified with the final temperature, NOT the Δt.

Also, I did this problem with 4.18. Doing it with 4.184 gives slightly different numbers. Make sure you check with your teacher as to the values of the various constants that he/she wishes for you to use.

Problem #2: Determine the final temperature when 18.0 g of ice at −10.0 °C mixes with 275.0 grams of water at 60.0 °C.

This is like problem 8e and several following in Worksheet #2.

The «cold water» in this case is going to do three things:

a) as a solid, warm up from −10 to zero
b) all 18 g will melt
c) as a liquid, the temperature goes up some unknown amount

The warmer water must provide all the energy for this to happen. Keeping in mind that the heat lost by the warmer must equal the heat gained by the colder (q_lost = q_gain), we have this:

heat to warm ice 10 degrees + heat to melt ice + heat to warm cold water by unknown amount = heat lost by the warm water

Here are the numbers:

Notice I used 6020 J rather than 6.02 kJ for the molar heat of fusion. That is because the other two parts of the left-hand side of the equation will give Joules as their answer. I used 6020 so that all three parts would be in Joules. If I had used 6.02, then that middle part woud have been in units of kJ.

Notice the use of x − 0 and 60.0 − x. This time, visualize (or write out) the number line used above. The zero is to the left, the 60.0 to the right and the x is in between the 0 and the 60.0.

Proceeding with the solution, I get:

and so, x = 51.3 °C.

Keep in mind that ‘x’ was identified with the final temperature, NOT the Δt.

Problem #3: 55.0 mL of ethanol (d = 0.789 g/mL) at 8.0 °C is mixed with 55.0 mL of water at 28.2 °C. Assuming no heat is lost, what is the final temperature of the mixture?

Two things must be done first: (1) determine how many grams of ethanol are present and (2) determine the specific heat for ethanol.

55.0 mL times the density (0.789 g/mL) gives the mass of ethanol present. This value is 43.395 g. I will use this value and round off at the end of the calculation.

Letting ‘x’ equal the final temperature, we have this:

You may finish this problem.

Problem #4: 900.0 L of palm oil at 47.0 °C is mixed with 200.0 L of palm oil at 76.0 °C. What is the final temperature? Assume no heat is lost to the surroundings.

1) Let us convert 900.0 L and 200.0 L to grams using the density:

(900,000 mL) (0.9 g/cm 3 ) = 8.10 x 10 5 g

(200,000 mL) (0.9 g/cm 3 ) = 1.80 x 10 5 g

2) We will let ‘x’ equal the final temperature, so the two temperature changes are:

the 900 L goes up in temperature, so use ‘x − 47.0’

the 200 L goes down in temperature, so use ‘76.0 − x’

3) Keeping in mind that the heat lost by the 200 L equals the heat gained by the 900 L, we write:

(1.80 x 10 5 ) (76.0 − x) = (8.10 x 10 5 ) (x − 47.0)

I did not write the specific heat values because they would just cancel out. This is because we have palm oil on both sides of the equation. If the substances were different (see examples above), the two specific heats would have to be included.

Also, when I did this problem on the calculator, I dropped the 10 5 portion of the mass. In addition, if you were to ignore the use of the density and use, say 200 and 900, you’d get 51.0 °C for the answer.

Problem #5: 364 g of water at 34.0 °C is added to ice at 0.0 °C. If the final temperature of the system (which you can assume is isolated) is 0.0 °C, determine how much ice melted. The specific heat of water is 4186 J/kg ⋅ °C. The latent heat of fusion for water is 335,000 J/kg.

1) Determine energy lost by warm water:

q = (0.364 kg) (34.0 °C) (4186 J/kg ⋅ °C)

2) Determine how much ice is melted by 51805.936 J:

51805.936 J / 335,000 J/kg = 0.155 kg = 155 g (to three sig figs)

Problem #6: How much ice (in grams) would have to melt to lower the temperature of 353.0 mL of water from 26.0 °C to 6.0 °C?

1) Determine heat lost by cooling water from 26.0 to 6.0:

Note the silent conversion from volume of water to mass of water using the density of 1.00 g/mL.

2) The heat lost by the warm water does two things:

a) melt an unknown mass of ice
b) take the same mass of melted ice from zero Celsius to 6 °C

3) We can express it thusly:

29539.04 J = (6020 J/mol) (x / 18.015 g/mol) + (x) (6.0 °C) (4.184 J g¯ 1 °C¯ 1 )

29539.04 J = 334.166x + 25.104x

Problem #7: An unknown volume of water at 18.2 °C is added to 27.8 mL of water at 33.6 °C. If the final temperature is 23.5 °C, what was the unknown volume? (Assume that no heat is lost to the surroundings; density of water is 1.00 g/mL.)

Note the our ‘x’ in this problem will be an unknown mass of water. Since we know the starting temperatures and the final temperature, we can calculate the Δt values.

1) Calculate the two Δt values:

cooler water ⇒ 23.5 − 18.2 = 5.2
warmer water ⇒ 33.6 − 23.5 = 10.1

This is 53.0 mL of water.

Problem #8: A student mixed 6.00 g of ice at −3.4 °C with 1.00 g of steam at 103.0 °C. What is the final temperature of this mixture?

heat gained in warming up = heat lost in cooling down

heat gained by ice + heat to melt ice + heat to raise water temp = heat lost by steam + heat lost as steam condenses + heat lost as water cools

(6.00 g) (3.4 °C) (2.06 J/g °C) + (6.00 g / 18.0 g/mol) (6020 J/mol) + (6.00 g) (x − 0 °C) (4.184 J/g °C) = (1.00 g) (3.0 °C) (2.02 J/g °C) + (1.00 g / 18.0 g/mol) (40700 J/mol) + (1.00 g) (100 − x °C) (4.184 J/g °C)

42.024 + 2006.667 + 25.104x = 6.06 + 2261.11 + 418.4 − 4.184x

Problem #9: A cube of ice is taken from the freezer at −5.5 °C and placed in a 98.0 g aluminum calorimeter filled with 326.0 g of water at room temperature of 20.0 °C. The final situation is observed to be all water at 15.0 °C. What was the mass of the ice cube?

1) The ice does three things:

(a) heat up from −5.5 to 0
(b) melt at 0
(c) heat up from 0 to 15

2) The three calculations are:

3) The water and calorimeter do this:

3) The two calculations are:

4) The amount of heat absorbed by the ice is equal to the heat lost by the water and Al calorimeter:

However, something important must be done and it’s associated with q_b.

(x) (5.5 °C) (2.02 J / g °C) + (x / 18.015 g/mol) (6020 J/mol) + (x) (15.0 °C) (4.184 J / g °C) = (326.0 g) (5.0 °C) (4.184 J / g °C) + (98.0 g) (5.0 °C) (0.900 J / g °C)

I changed 6.02 kJ/mol to 6020 J/mol.

11.11x + 334.166x + 62.76x = 6819.92 + 441

Problem #10: 20.0 g of steam at 100.0 °C is bubbled into a mixture of 50.0 g of water and 200.0 g of ice at exactly 0 °C. All of the steam condenses to water. What is the composition of the system at the end?

Latent heat of fusion of ice = 334.16 J/g
Latent heat of vaporization of water = 2259.2 J/g

1) Let’s condense the 20.0 g of steam and then cool it to zero:

(20.0 g) (2259.2 J/) = 45184 J
(20.0 g) (100 °C) (4.184 J/g °C) = 8368 J

45184 + 8368 = 53552 J

2) We now have a system of 70.0 g of water at 0 °C and 200.0 g of ice with 53552 J «floating» about. (The 70.0 comes from the 50.0 already there plus the 20.0 of the steam that condensed.) Let’s see how much ice will be melted by the 53552 J:

3) How much ice is left?

1) The ice will absorb energy from the warm water. That energy gained makes the ice do three things:

1) heat from −6.8 °C to 0 °C
2) melt at 0 °C 3) heat from 0 °C to the final temperature

2) Here the equations for the three energy-gaining behaviors of the ice:

q₁ = (35.46 g / 18.0 g/mol) (6.8 °C) (37.5 J K¯ 1 mol¯ 1 ) 1 mol¯ 1 ) 1 mol¯ 1 )

4) You solve this equation:

Note that only one variable has been used. The ‘x’ is the final temperature.

5) Substitute into the above and do a bit of algebra.

[(1.97) (6.8) (37.5)] + [(1.97) (6010)] + [(1.97) (x) (75.3)] = (3.84) (91.0 − x) (75.3)

502.35 + 11839.7 + 148.341x = 26312.832 − 289.152x

1) I changed 6.01 kJ/mol to 6010 J/mol so as to align all the units to Joules.
2) I ignored the °C and K issues on the temps. That’s because all the temps involved are changes in temp, not a single, fixed temperature. Since 1 °C is the same size as 1 K, everything cancels.

Problem #12: A 40.0 g ice cube at −19.0 °C are placed into 275 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts.

Heat capacity of H₂O(s) = 37.7 J/(mol K)
Heat capacity of H₂O(ℓ) = 75.3 J/(mol K)
Enthalpy of fusion of H₂O = 6.02 kJ/mol

1) The ice cube does three things:

(a) heat from −19 to zero C
(b) melt at zero C
(c) heat up from zero to the final temp

2) The liquid water does one thing. It cools down from 25.0 °C to the final temp.

3) The heat gained by the ice equals the heat lost by the 25.0 °C of water.

4) The energy involved in the ice cube:

q_a = (40.0 g / 18.0 g/mol) (19 K) (37.7 J/(mol K))
q_b = (40.0 g / 18.0 g/mol) (6020 J/mol)

1) The key assumption is that all of the heat lost by the warmer water is gained by the cooler alcohol. In an equation:

(mass of water) (temp change of water) (specific heat of water) = (mass of alcohol) (temp change of alcohol) (specific heat of alcohol)

2) Energy lost by the water:

(0.300 kg) (10.0 °C) (4.184 kJ kg¯ 1 °C¯ 1 ) = 12.552 kJ

3) 12.552 kJ will be gained by the alcohol:

x = 3.1375 kJ kg¯ 1 °C¯ 1

To three significant figures, the answer is 3.14 kJ kg¯ 1 °C¯ 1

4) Often you will find steps 2 and 3 above combined as follows:

(0.300 kg) (10.0 °C) (4.184 kJ kg¯ 1 °C¯ 1 ) = (0.200 kg) (20.0 °C) (x)

1) The energy from the warm water will do two things:

(a) heat up the cooler water from 23.6 °C to 38.9 °C
(b) heat up the calorimeter from 23.6 °C to 38.9 °C

2) In so doing, the temperature of the warmer water will drop from 73.3 °C to 38.9 °C. Since the warmer water is the only source of heat energy, let up determine how much heat is lost:

q = (41.58 g) (34.4 °C) (4.184 J g¯ 1 °C¯ 1 ) = 5984.59 J

3) From step 1 above, here are the set-ups for (a) and (b):

4) The sum of (a) and (b) equals 5984.59 J:

5984.59 J = (m) (15.3 °C) (4.184 J g¯ 1 °C¯ 1 ) + 402.39 J

m = 87.2 g (to three sig figs)

Problem #15: Equal masses of hot water and ice are mixed together. All of the ice melts and the final temperature of the mixture is 0 °C. If the ice was originally at 0 °C, what was the initial temperature of the hot water?

1) Let us assume we have 18.0 g of ice and 18.0 g of hot water present.

2) The key is to realize that the only thing the ice did is melt, it did not change its temperature. So, let us calculate the amount of heat needed to melt our 18.0 g (or, 1.00 mole) of ice:

3) The only source of heat is the hot water, which provides 6020 J (I converted the 6.02 kJ to J.) of heat. Let us calculate the temperature change of 18.0 g of hot water as it loses 6020 J of heat:

The hot water was at an initial temperature of 79.9 °C and went to 0 °C as the ice melted (and stayed at 0 °C the entire time).

Problem #16: In real calorimeters, most of the heat released by the bomb is absorbed by water, but a certain amount is also absorbed by the metal and insulation surrounding the water tank. A certain calorimeter absorbs 24 J/°C. If 50.0 g of 52.7 °C water is mixed with the calorimeter’s original 50.0 g of 22.3 °C water, what will be the final temperature of the mixture?

1) We have some warm water that will lose heat. We can set up a calculation to determine the amount:

The T is the final temp and (52.7 °C − T) is the temp change.

2) All the heat that is lost goes two places:

(a) the cool water
(b) the calorimeter itself

3) Here’s the set up for (a):

Remember, the T (final temp) is higher than 22.3 °C so the temp change starts at 22.3 °C and goes up to T.

4) Here’s the set up for (b):

The calorimeter is at the same starting temp (22.3 °C) as the cool water and it winds up at the same T (final temp) so it goes through the same change in temp (called the ΔT)

(50.0 g) (4.184 J/g °C) (52.7 − T) = (50.0 g) (4.184 J/g °C) (T − 22.3) + (24 J/°C) (T − 22.3)

The left side is the heat coming out of the warm water and the right side is the heat going into two destinations (the calorimeter itself and the cool water).

11024.84 − 209.2T = (209.2T − 4665.16) + (24T − 535.2)

11024.84 − 209.2T = 233.2T − 5200.36

T = 36.7 °C (to three sig figs)

Problem #17: 40.0 g of ice at 0.0 В°C is combined with 150. g of liquid water at 80.0 °C in a coffee cup calorimeter. Calculate the final temperature reached, assuming no heat loss or gain from the surroundings.

We know the liquid water will come down in temperature, perhaps all the way to zero. We also know that some (if not all) of the ice will melt. I’m going to see how much energy is required to melt all the ice and then see how that compares to the energy given off as the liquid water cools down.

1) How much energy is required to melt all the ice:

(40.0 g / 18.015 g/mol) (6.02 kJ/mol) = 13.367 kJ = 13,367 J

2) How much energy is liberated by the liquid water as it cools down to zero Celsius:

q = (150. g) (80.0 °C) (4.184 J/g °C)

This calculation shows that there is sufficient energy to melt all the ice. Sometimes, the teacher will give you a problem where not all the ice is melted by the energy from the warm water. In that case, the final temperature would be zero Celsius. See below for an example.

3) We need to know what amount of temperature change happens when 13,367 J of energy is removed from the warm water.

13367 J = (150. g) (Δt) (4.184 J/g °C)

Δt = 21.2986 °C (I’ll round off the final temp to three sig figs.)

4) We now have a new problem: «What is the final temperature when 40.0 g of water at 0.00 °C is mixed with 150. g of water at 58.7014 °C.»

(40.0 g) (x − 0 °C) (4.184 J/g °C) = (150. g) (58.7014 °C − x) (4.184 J/g °C)

(40.0 g) (x − 0 °C) = (150. g) (58.7014 °C − x)

40x = 8805.21 − 150x

Problem #18: 40.0 g of ice at 0.0 В°C is combined with 150. g of liquid water at 20.0 °C in a coffee cup calorimeter. Calculate the final temperature reached, assuming no heat loss or gain from the surroundings.

1) How much energy is required to melt all the ice:

(40.0 g / 18.015 g/mol) (6.02 kJ/mol) = 13.367 kJ = 13,367 J

2) How much energy is liberated by the liquid water as it cools down to zero Celsius:

q = (150. g) (20.0 °C) (4.184 J/g °C)

3) Some discussion:

This calculation shows that there is insufficient energy to melt all the ice. That means that the final temperature is zero Celsius and there is some unmelted ice floating in the liquid water.

At this point, the problem could ask how much ice remains unmelted. The answer for the above problem is 2.44 g. I will leave you to ponder how to figure it out.

Problem #18: Equal masses of liquid A, initially at 100 °C, and liquid B, initially at 50 °C, are combined in an insulated container. The final temperature of the mixture is 80 °C. Which has the larger specific heat capacity, A or B?

The heat given off by A is equal to the heat absorbed by B. No heat is lost to the container.

2) In an equation, we write this:

3) This, therefore, is also true:

(mass of A) (temp change of A) (sp heat of A) = (mass of B) (temp change of B) (sp heat of B)

4) The masses are equal, giving:

(temp change of A) (sp heat of A) = (temp change of B) (sp heat of B)

The specific heat capacity of A is larger than that of B.

Bonus Problem: Water at 0 °C was placed in a dish inside a vessel maintained at low pressure by a vacuum pump. After a quantity of water had evaporated, the reminder froze. If 9.31 g of ice at 0 °C was obtained, how much liquid water must have evaporated? (At 0 °C, ΔH_fusion = 6.02 kJ/mol; ΔH_vap = 45.054 kJ/mol)

1) The water that froze (9.31 g) released:

(9.31 g) (6.02 kJ/mole) (1 mole / 18.015 g) = 3.11108 kJ

2) This energy is given up by the water vaporizing:

energy = (moles of water that evaporated) (heat of vaporization)

3.1108 kJ = (mass of water that evaporates) (45.054 kJ/mole)

0.069046 moles of water evaporated

(0.069046 mol) (18.015 g/mol) = 1.24 g (to three sig figs)

Note the use of ΔH_vap for 0 °C. At 100 °C, it equals 40.7 kJ/mol. (Here is the source I used for the 45.054 value.)

Finding the final temperature of a mixture of ice and steam

Answers and Replies

Using which formula? The equation in post #1 has a sign error, as I indicated.
Please post the equation you are now using and the subsequent working.
I get quite a bit more than 12C.

To get a feel for what the answer ought to be, note that although there is 5 times the mass of ice the latent heat of vaporisation is about seven times that of fusion, so it should be a middling sort of number.

Using which formula? The equation in post #1 has a sign error, as I indicated.
Please post the equation you are now using and the subsequent working.
I get quite a bit more than 12C.

To get a feel for what the answer ought to be, note that although there is 5 times the mass of ice the latent heat of vaporisation is about seven times that of fusion, so it should be a middling sort of number.

Your original equation
##Q_1 + Q_2 = Q_3 + Q_4##
is based on ##Q_1, Q_2, Q_3## and ##Q_4## all being positive quantities.

##Q_4## is the thermal energy released when the water from the (condensed) steam cools from ##T_i = 100^oC## to ##T_f##.

«Q3 is the amount of heat required to condensate the steam»
No! No energy is required (i.e. you don’t need to supply energy).

Have you ever boiled water to make a cup of tea? You *require* (must supply) an amount of energy ##ml_v## to convert water to steam.

But you do not need to supply any energy to condense the steam to water. In fact the steam *releases* an amount of energy ##ml_v## when it condenses.

In words, the equation ##Q_1 + Q_2 = Q_3 + Q_4## means:
Het required to melt ice
+ Heat required to raise ice from ##0^oC## to ##T_f##
=
Heat released when steam condenses
+
Heat released by condensed steam cooling from ##100^oC## to ##T_f##

Expressed this way, ##Q_1,Q_2, Q_3## and ##Q_4## must all be positive quanties.