Temperature Equilibrium

  • Thread starter MetalCut
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  • #1
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The question is:
When you take a bath, how many kilograms of hot water(49.0 Degrees Celcius) must you mix with cold water(13.0 Degrees Celcius) so that the temperature of the bath is 36.0 Degrees Celcius. The total mass of water(hot + cold) is 191kg. Ignore any heat flow between the water and the external surroundings.

As far as i know Q=mL, where m is the mass of the cold water,
Q=cm(change in T), where T is the higher temperature minus the lower eauilibrium temperature. And that the equilibrium temperature is 36.

But i don't understand how to deal with this question if the temperature given is for both hot and cold water. If i had one of those, the question would have been easy.:cry:
 

Answers and Replies

  • #2
Andrew Mason
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MetalCut said:
The question is:
When you take a bath, how many kilograms of hot water(49.0 Degrees Celcius) must you mix with cold water(13.0 Degrees Celcius) so that the temperature of the bath is 36.0 Degrees Celcius. The total mass of water(hot + cold) is 191kg. Ignore any heat flow between the water and the external surroundings.

As far as i know Q=mL, where m is the mass of the cold water,
Q=cm(change in T), where T is the higher temperature minus the lower eauilibrium temperature. And that the equilibrium temperature is 36.

But i don't understand how to deal with this question if the temperature given is for both hot and cold water. If i had one of those, the question would have been easy.:cry:
You know that there is a loss of 13 degrees of the hot water and a gain of 23 degrees for the cold. Since heat energy transferred/unit mass is proportional to temperature change, what does this tell you about the relative proportion of hot to cold water initially?

AM
 
  • #3
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Yes I know that, but still no idea........
Is there like a ratio or something.
 
  • #4
Andrew Mason
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MetalCut said:
Yes I know that, but still no idea........
Is there like a ratio or something.
Let x represent the mass of cold water and y the mass of hot water.

We know that x + y = 191.

We also know that the heat flowing gained by the cold is equal to the heat lost by the hot. That heat is [itex]Q = m\Delta T[/itex]

So: Q = 23x = 13y --> x = 13y/23

Work that out to find x and y.

AM
 
  • #5
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Correct me if i'm wrong, but can i not do it like this:

The cold water went from 13-36, so 23 is the difference.
23/36*(100) = 63.8%
So then 63.8% of 191kg is 121kg. That would then be the mass of the hot water.
The hot water went from 49-36, so 13 is the difference.
13/36*(100) = 36.2%
And then 36.2% of 191kg is 70kg. That would be the mass of the cold water.

Is this right or is it a coincidence?
 
  • #6
Andrew Mason
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MetalCut said:
Correct me if i'm wrong, but can i not do it like this:

The cold water went from 13-36, so 23 is the difference.
23/36*(100) = 63.8%
So then 63.8% of 191kg is 121kg. That would then be the mass of the hot water.
The hot water went from 49-36, so 13 is the difference.
13/36*(100) = 36.2%
And then 36.2% of 191kg is 70kg. That would be the mass of the cold water.

Is this right or is it a coincidence?
It is a coincidence. The answer is about right but the reasoning is not. You are using 36 because it is the final temperature. Why would you divide by the final temperature? What if the temperatures were in degrees K?

The only reason it works out is because, coincidentally, the sum of the temperature differences is equal to the final temperature in C.

Use simple algebra:

(1) x + y = 191 and
(2) 23x = 13y

Substitute x = .57y from (2) into (1) to get 1.57y=191 so y = 122 kg. and x = 69 kg.

AM
 
  • #7
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Ok thanx man. Appreciate it. C ya.
 

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