Thermodynamics (final temperature, ice mixed with steam)

[SigFig]

Homework Statement

If 10-kg of ice at 0 degrees Celsius is added to 2-kg of steam at 100 degrees Celsius, the temperature of the resulting mixture is? Use Joules. ANSWER: 40

Homework Equations

Specific Heats:
Ice = 2060
Steam = 2020

Latent Heats:
Water melting = 3.33x10^5
Water boiling = 2.26x10^6

The Attempt at a Solution

Q1=mL
Q1=10(3.34x10^5)
Q1=3.34x10^6

Q2=(2)(2.26x10^6)
Q2=4.52x10^6

Q2-Q1=1.18x10^6

Q=MCΔT
1.18x10^6 = (2+10)(4180)(T-0)
1.18x10^6 = 5.0x10^4T
T=23.6

Sadly 23.6 =/= 40
I've been trying to do this problem for over an hour now using every method I remember, stumped.

 

[JesseC]

I'm confused by the question. Steam is a suspension of water droplets in air, basically a cloud. Presumably you mean water vapour not steam?

In this system there will a phase change from ice to water that costs an amount of latent heat. There will also be a phase change from vapour to liquid, which releases an amount of latent heat.

Perhaps it is easier to think about in too stages. 1) The vapour releases its latent heat and becomes water. This heat is sufficient to provide the latent heat required to melt the ice to water at 0C, and then heat up the mass of water that was ice. So finally you're left with two blobs of water, 2kg at 100C and 10kg at some temperature above 0C.

The second stage (that you missed out): the temperature of the two water masses equilibrate.

 

[collinsmark]

Homework Statement

If 10-kg of ice at 0 degrees Celsius is added to 2-kg of steam at 100 degrees Celsius, the temperature of the resulting mixture is? Use Joules. ANSWER: 40

Homework Equations

Specific Heats:
Ice = 2060
Steam = 2020

Latent Heats:
Water melting = 3.33x10^5
Water boiling = 2.26x10^6

Don't forget your units.

The Attempt at a Solution

Q1=mL
Q1=10(3.34x10^5)
Q1=3.34x10^6

Q2=(2)(2.26x10^6)
Q2=4.52x10^6

Q2-Q1=1.18x10^6

This is correct so far, except for the lack of units.

Q=MCΔT
1.18x10^6 = (2+10)(4180)(T-0)

Your approach *would* be correct, *if* the entire 12 kg of [liquid] water started from 0 deg C, and increased its temperature from there. But that's not the case for this problem.

Here, there is only 10 kg of [liquid] water, starting at 0^o C, rising to temperature T, absorbing energy as it goes.

There is also 2 kg of [liquid] water, starting at 100^o C, falling to temperature T (making a temperature change of 100 - T), releasing energy as it goes.

You'll need to re-set up your equation to account for both of these things.

I'm confused by the question. Steam is a suspension of water droplets in air, basically a cloud. Presumably you mean water vapour not steam?

That's really a matter of semantics. But commonly, water in gaseous form is called steam. (In which case -- you are correct insofar that -- H₂O is transparent when in truly gaseous form.)

 

[SteamKing]

It would also be nice if some units were included in the calculations.

 

[Nickie]

As a scientist, it is important to check for any errors in the calculations and make sure all units are consistent. In this case, there seems to be an error in the calculation for Q1, the heat absorbed by the ice. The correct calculation should be Q1 = 10(2060)(T-0) = 20,600T. This would result in a different final temperature, T = 36.2 degrees Celsius. Therefore, the answer of 40 is not correct and further review and correction of the calculations is needed. Additionally, it is important to note that this calculation assumes ideal conditions and does not account for any heat loss or gain from the surroundings.