# Heat Transfer from water to ice and calculation of the final temperature

I'm real confused on ow to do this one. I have another just like it. I know you have to use a couple of equations --q=m x c x change in temp. and the Hf=m x heat of fusion

but i'm not sure how to get to my answer. Thanks for the help!!!

What is the final temperature, (in oC), after a 21.5 gram piece of ice at 0oC is placed into a styrofoam cup with 119.0 grams of water initially at 74.0oC? Assume no loss or gain of heat from the surroundings. Enter your answer without units. heat of fusion of water is 333 J/g. The specific heat of H2O(l) is approximately constant at 4.184 J/gK.

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The big idea in calorimetry is that the total of all energy changes must be zero (that is, energy can not be created nor destroyed). So,

energy absorbed to melt ice + energy given off as liq water cools = 0

Rearranging yields:

energy absorbed to melt ice = - energy given off as water cools *

The equation which describes the energy absorbed to melt ice is

energy to melt ice = mass of ice x heat of fusion (333 J/g)

The equation that describes the energy given off by the water as it cools is

energy given off by cooling = mass of liquid water x (Temp final - Temp initial) x specific heat (4.184J/g)

By substituting in the original equation (*) we get

mass of ice x Heat of fusion = mass of liquid water x (T final - T initial) x sp. heat

All values are given with the exception of the final temperature. Plug in values and solve for the final temperature.

(Note this solution is using a simplification, if the computer doesn't agree, we can talk about a more exact solution)

spock

Mon, 2008-05-12 19:19

Ok i got it wrong. it said to

Set up an equation to balance the heats: the heat to melt the ice and to warm its cold water must equal the heat used to cool the hot water to the final temperature.

My work using your equations from above

(21.5 g ice) x (333 J/g(= (18.01 g H2O)( x (X)(or 74 C??) x (4.184 j/gK)

= (7159.5)=(75.353) x (X)

7159.5/75.353=95 C but its incorect.

I oviously messed up. But where!!

antgoblue

Mon, 2008-05-12 21:46

Well, I wasn't sure how much approximation they were willing to accept. But, your post answers that question.

I did NOT include the heat absorbed by the ice water after it melts as it warms up to the final temperature. In addition, it would appear that you made several errors in plugging into the formula.

So first the corrected formula:

energy absorbed to melt ice + energy absorbed to warm ice water + energy given off as liq water cools = 0

Rearranging the equation yields:

energy absorbed to melt ice + energy absorbed to warm ice water = - energy given of as liq water cools

mass of ice x Heat of Fusion + mass of ice x (Tfinal - Tinitial) x sp heat = -(mass of liq water x (Tfinal - Tinitial) x sp. heat

Keep in mind that there will only be one final temperature. The melted ice water will warm up and the warm water will cool down until they come to a common temperature. So the final temperature of both of them can be represented by Tfinal. The ice water will begin warming from an initial temperature of 0C while the warm water will begin cooling from an original temperature of 74C.

Plugging in yields:

( 21.5g x 333J/g ) + 21.5g x (Tfinal - 0C) x 4.184J/g C) = - (119.0g x (Tfinal - 74C) x 4.184J/gC)

( 7160 J ) + (29.8 x Tfinal) = - (497.9 Tfinal - 36,800)

Multiply through on the right side by the - :

( 7160 J ) + (29.8 Tfinal)J = - ( 497.9 Tfinal + 36,800)J

Get the Tfinal on the left and the other numbers on the right:

29.8 Tfinal + 497.9 Tfinal = 36,800 -7160

(29.8 + 497.9) Tfinal = 29640

527.5 Tfinal = 29640

Tfinal = 29640 / 527.5

Tfinal = 56.2 C

spock

Mon, 2008-05-12 22:36

Thanks Sock but I put in your answwer and the program says its wrong. Do you see a mistake somewhere?

antgoblue

Tue, 2008-05-13 16:23

The setup is correct:

( 21.5g x 333J/g ) + 21.5g x (Tfinal - 0C) x 4.184J/g C) = - (119.0g x (Tfinal - 74C) x 4.184J/gC)

There is a math/calculator error: 21.5 x 4.184 is not equal to 29.8 (in red)

( 7160 J ) + (29.8 x Tfinal) = - (497.9 Tfinal - 36,800)

I would suggest that you correct my error and then work through the rest of the calculations using my previous post as a guide and see if you can't come up with the correct answer.

spock

Tue, 2008-05-13 17:10

Yeah Spock I got it correct but got this one wrong!!

What amount of energy in Joules is necessary to melt a 19.5 gram piece of ice at 0 to form liquid water at the same temperature 0? heat of fusion of water is 6.01 kJ/mol or 333 J/g The specific heat of H2O(l) is approximately constant at 4.18 J/g*C. Enter your answer in scientific notation with no units.

you use q=m*c*change in temp. and mHf for heat of fusion

c=4.18J/g*C

change in temp = 0

m=19.5 g ice

heat fusion=333J/g

heat of fusion = 19.5 g * 333J/g=6493.5 J

energy in J= 19.5g * 4.184J/gC * 0 degrees (or 1???)=81.588 J

6493.5J/81.588J=79.58 J and this incorrect! where did I mess up! Thanks!

antgoblue

Tue, 2008-05-13 17:54

It would appear that this is not the same problem as the first one. They seem to be asking how much energy would be required to melt a 19.5 piece of ice at the melting point. Since the temperature of the ice/water is not changing, but staying at 0C, the ONLY flow of heat that we are interested in is into the ice cube. The heat required to melt the ice would be given by this equation:

heat to melt ice = mass of ice x heat of fusion

= 19.5 g x 333 J/g

= 6,494 J = 6.494 x 10^3 J

spock

Tue, 2008-05-13 19:25

Ok that ones done. Once again heres another. I also need help on the one above I can't get that one to save my soul!!

Icebergs in the North Atlantic present hazards to shipping, causing the length of shipping routes to increase by about 30 percent during the iceberg season. Attempts to destroy icebergs include planting explosives, bombing, torpedoing, shelling, ramming, and painting with lampblack. Suppose that direct melting of the iceberg, by placing heat sources in the ice, is tried. How much heat is required to melt 15 percent of a 1.00×105 metric-ton iceberg? One metric ton is equal to 103 kg. Assume that the iceberg is at 0°C. (Note: To appreciate the magnitude of this energy, compare your answer to the Hiroshima atomic bomb which had an energy equivalent to about 15,000 tons of TNT, representing an energy of about 6.0×1013 J.)

All these #'s are super confusing!!

antgoblue

Tue, 2008-05-13 19:31

You need to read the question carefully and pull out the important part:

"How much heat is required to melt 15 percent of a 1.00×105 metric-ton iceberg? One metric ton is equal to 103 kg. Assume that the iceberg is at 0°C. One metric ton is equal to 103 kg."

You need to find the heat to melt ice, that depends on only 2 things: the mass of the ice and the heat of fusion.

So, you need to use the information to find the mass of ice in grams:

mass of ice = .15 x 1.0 x 10^5 metric tons ice x 10^3 kg/ton x 10^3 grams/kg = ?? grams of ice

Once you solve for the grams of ice, you plug it into the heat of fusion equation:

heat = mass of ice x heat of fusion (we have used the value for the heat of fusion in earlier problems)

spock

Tue, 2008-05-13 21:46

ok so its:

(.15) x (1.0 x 10^5) x (10^3) x (10^3) x (10^3) = 1.5 x 10^10

heat= mass ice x heat fusion

= 1.5 x10^10 x 4.184 = 6.276 x 10^10 and its wrong

Where did i mess up?

antgoblue

Tue, 2008-05-13 22:03

specific heat is 4.184 J/g, the heat of fusion is 333 J/g.

spock

Tue, 2008-05-13 22:07

Got it!!! One more...

Ethanol boils at a temperature of 78.29oC. What amount of energy, in joules, is necessary to heat to boiling and then completely vaporize a 18.0 gram sample of ethanol initially at 11oC? The ?Hvap is 38.56kJ/mol and the specific heat of ethanol(l) is approximately constant at 2.44 J/gK. The formula for ethanol is C2H5OH. Enter your answer in scientific notation without units.

antgoblue

Tue, 2008-05-13 22:13

You need to be doing these! I've showed you how to do a number of them, now you should be at least trying them on your own.

Boiling works just like melting accept you use the heat of vaporization rather than the heat of fusion

heat absorbed in melting = mass x heat of fustion

heat absorbed in boiling = mass x heat of vaporization.

spock

Tue, 2008-05-13 22:38

I tied it and got it right. I can't get one of them. Its exactly like the 1st problem we did. I am confused on the setup and where to go from there!!

What is the final temperature, (in ), after a 19.5 gram piece of ice at 0 is placed into a styrofoam cup with 116.0 grams of water initially at 78.5? The mass of the ice, and the temperature and mass of the hot water are the same as in the previous two questions. Assume no loss or gain of heat from the surroundings. Enter your answer in scientific notation without units.

I did:

( 19.5g x 333J/g ) + 19.5g x (Tfinal - 0C) x 4.184J/g C) = - (116.0g x (Tfinal - 78.5C) x 4.184J/gC)

= 38099.504=27250.392

38099.504/27250.392=1.39 C and its inccorect! Thanks Again!!!!!!!!!

antgoblue

Wed, 2008-05-14 18:35

I still cannot get it right!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

antgoblue

Wed, 2008-05-14 20:27

Help me please!!

antgoblue

Wed, 2008-05-14 22:29

( 19.5g x 333J/g ) + 19.5g x (Tfinal - 0C) x 4.184J/g C) = - (116.0g x (Tfinal - 78.5C) x 4.184J/gC)

I just did a quick glance at your setup, but from that glance, you have T

_{final}on both sides of the equation. You have to combine like terms (T_{final}) first before you solve for the variable.valdorod

Wed, 2008-05-14 22:37

ok so i changed it around and got 9.9 as an answer and this is correct. Where did i mess up now!!

antgoblue

Wed, 2008-05-14 22:53

Your set up is fine:

( 19.5g x 333J/g ) + 19.5g x (Tfinal - 0C) x 4.184J/g C) = - (116.0g x (Tfinal - 78.5C) x 4.184J/gC)

Your math after that point is not correct. You can't just drop Tfinal out of the equation!

What is 19.5g x (Tfinal - 0C) x 4.184J/g C) equal to?

What is (116.0g x (Tfinal - 78.5C) x 4.184J/gC) equal to?

(Hint: a (x - b)c = acx -acb )

spock

Wed, 2008-05-14 23:25

As you said before it is just like the first problem, spock worked out the first problem for you

Follow the same procedure, just replace 21.5 g with 19.5 g for the ice and

replace 119.0 g with 116 g.

Notice that both masses are decreased only slightly, you final answer should be a few degrees different from the first example (hint it should be in the 50's)

Notice how spock ended up with T

_{final}on both sides of the equationThen he collected like terms, combined both T finals on one side and the energies on the other side. then divided.

valdorod

Wed, 2008-05-14 23:28

Yeah i got the like terms together and got the right answer! Thanks for the help!

antgoblue

Thu, 2008-05-15 17:33