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Old Oct 16, 2006 | 06:06 PM
  #16  
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Do you just have to find the temperature of the water when you started, or do you have to be able to measure the temperature immediately?

For example, can you let the water cool off for an hour and then say what the temperature was before it cooled off?

You could probably use Newton's law of cooling and work back to your initial temperature if you knew room temperature, and measured final temperature after 20 minutes or so. (Plus get one more measurement at like 25 minutes to determine the experimental constant in the eqaution)
 
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Old Oct 16, 2006 | 06:07 PM
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Could you please write me the equation?
I understand the steps; however, I don't understand what to do at the end, after all measurements were taken. How do I find original temperature of water?
 
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Old Oct 16, 2006 | 06:07 PM
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Originally Posted by isubsmith
Do you just have to find the temperature of the water when you started, or do you have to be able to measure the temperature immediately?

For example, can you let the water cool off for an hour and then say what the temperature was before it cooled off?

You could probably use Newton's law of cooling and work back to your initial temperature if you knew room temperature, and measured final temperature after 20 minutes or so. (Plus get one more measurement at like 25 minutes to determine the experimental constant in the eqaution)
What is exactly what I need to do.
 
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Old Oct 16, 2006 | 06:08 PM
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Originally Posted by andym
Ithe original temp would be easy to figure out from there.
Ok, if you have 1L of 100C water, and 1L of 50C water, and mix the two of them in a 2L container, it's going to be 75C, minus a little bit because the container is going to drain a little heat, and minus another little bit because it took you a minute to pour both smaller container in without spilling any.

I don't know how real-world the answer is supposed to be. If you ignore air resistance, gravity, and a whole lot of other things that make real-world physics nearly impossible to figure out, it becomes a lot easier.
 
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Old Oct 16, 2006 | 06:10 PM
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Originally Posted by andym
Ok, if you have 1L of 100C water, and 1L of 50C water, and mix the two of them in a 2L container, it's going to be 75C, minus a little bit because the container is going to drain a little heat, and minus another little bit because it took you a minute to pour both smaller container in without spilling any.

I don't know how real-world the answer is supposed to be. If you ignore air resistance, gravity, and a whole lot of other things that make real-world physics nearly impossible to figure out, it becomes a lot easier.
But I don't know at what temperature water is. That's what I need to find out.
 
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Old Oct 16, 2006 | 06:14 PM
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That was a theorhetical example to show you how to calculate the temperature.

Instead of essentially averaging the two temps (100 and 50 in the above example) you have to work backwards. You have the starting temp of one and the result - you need to calculate the other starting temp.

I'm showing you that 5 + 3 = X, and you have to take that and turn it into 5 + X = 8 and solve for X.
 
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Old Oct 16, 2006 | 06:16 PM
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No, I'm business major.

This would be a lot easier if you write me the equation.
 
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Old Oct 16, 2006 | 06:19 PM
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Originally Posted by Red Star
But I don't know at what temperature water is. That's what I need to find out.
Say you have 3L of 25C water and 1L of unknown temperature water. Pour that together and get your final temperature.

Now you have (25+25+25+X)/4 == Final Temperature. Solve for X there.
 
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Old Oct 16, 2006 | 06:29 PM
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That looks way to simple to be college physics.
 
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Old Oct 16, 2006 | 06:30 PM
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Originally Posted by seventyseven250
OK, I think AndyM and I are on the same wavelength.

Do this:
- Get some cold water and measure its temperature. Meauser its volume as well.
- Measure the volume of warm water (of unknown temperature).
- Mix the two (in an insulated container so no heat is lost)
- Measure the resulting temperature. Since you know the volumes of both, plus the temp of one, plus I'm assuming you know the spcific heat of water in liquid form, you only have on unknown in the equation.

This assumes that the resulting temp will be within the range that your thermometer can measure.
Sound good?
This looks good.

Only thing that I need is equation.
 
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Old Oct 16, 2006 | 06:36 PM
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Use a candy thermometer

Celcius to Fahrenheit.

double it and add 30.

Ie:50*C x 2 +30 = 130*F
100*C x 2 +30 = 230*F

Now if you are talking centigrade and not celcius. that's anothe equation
 
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Old Oct 16, 2006 | 06:39 PM
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That's not what I need.

I need equation for steps that seventyseven250 wrote.
 
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Old Oct 16, 2006 | 08:02 PM
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Well, I don't know the answer to your question, but something I learned, the hard way, is that it doesn't pay to spend good money on tuition and not attend class.
 
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Old Oct 16, 2006 | 08:26 PM
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Originally Posted by scole250
Well, I don't know the answer to your question, but something I learned, the hard way, is that it doesn't pay to spend good money on tuition and not attend class.
Too True, unless it's a powder day, and I don't think that applies at this time of year.

If the equation isn't in your textbook, then I think you are hooped. I'm not digging out my Physics textbook for this one. Look up "specific heat" in the index and that should get you to the right section . . .
 
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