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45g of water at 4 degrees C is added to 150g of water at 75 degrees C. What is the mixture's equilibrium temperature?

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- Thread starter Tokimasa
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- #1

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45g of water at 4 degrees C is added to 150g of water at 75 degrees C. What is the mixture's equilibrium temperature?

- #2

FredGarvin

Science Advisor

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[tex]q = m*C_p *\Delta T[/tex]

Both objects will have the same heat transfer.............

Both objects will have the same heat transfer.............

- #3

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FredGarvin said:[tex]q = m*C_p *\Delta T[/tex]

Both objects will have the same heat transfer.............

I don't get it...

m = total mass

c = forgot it's name of water

delta T = unknown

q = unknown

You have two unknowns...

- #4

GCT

Science Advisor

Homework Helper

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q1 (gained)=-q2(lost), you'll need to set the q equal to each other and solve for final temperature, Tf, which will be the same for the two equations.

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- #6

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hey tokimasa

m*c*(T-277)=-M*c*(T-348)

we know m,M and we can cancel "c" because both c are equal(we are talking about mixing the same material here-water)...then we will get an equation based on T...and we can find T from it,which is the temperature of the mixture.......

now i will tell u the idea behind this......we are talking about mixing the same substance...that is water......so the property of heat transfer and temperature changes can be easily related.....when u bring cold water towards hotter water...the hot loses some energy...ie Q...this Q is gained by cold water...then only the condition becomes equillibrium isnt it???????????here one is loosing and the other is gaining...thats y the sign difference occurs

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- #7

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nomorevishnu said:hey tokimasa

m*c*(T-277)=-M*c*(T-348)

we know m,M and we can cancel "c" because both c are equal(we are talking about mixing the same material here-water)...then we will get an equation based on T...and we can find T from it,which is the temperature of the mixture.......

now i will tell u the idea behind this......we are talking about mixing the same substance...that is water......so the property of heat transfer and temperature changes can be easily related.....when u bring cold water towards hotter water...the hot loses some energy...ie Q...this Q is gained by cold water...then only the condition becomes equillibrium isnt it???????????here one is loosing and the other is gaining...thats y the sign difference occurs

I see where you are coming from now. But where do you get 227 and 348 from?

- #8

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The heat gained frmo one equals the heat lost from the other

[tex] q_1 = -q_2 [/tex]

[tex] m*c*(T_{celsius}+273) = -m*c*(T_{celsius} + 273) [/tex]

Where the left side is one of the liquids, and the right side is the other. C is the same for both, and m is the mass of each quantity.

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- #10

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But it is wrong.. T - 277? T - 348? Whats that?

- #11

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hey....Q=m*c*(Tf-Ti).......

in the problem given by tokimasa........Tf...the final equilibrium temperature is what we have to find......for both the part of liquids change of temperature takes place..........for the cooler water temperature increases from 277K(given by tokimasa) to Tf....so...Q=m*c*(Tf-277)

for the cooler water temperature decreases from 348K(also given) to Tf....

so Q=M*c*(Tf-348)

remember......delta T is always final-initial...so Tf-277 and Tf-348 likewise....now do u understand????????????????

- #12

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First of all, is it really necessary to change it to Kelvin? Is the answer meant to be in Kelvin?

It is quite simple. You find m*c*DT for both of the "waters". You know the mass of both, the specific heat capacity (c) is 4190, and you know the initial temperature of both systems. You want to find the final temperature.

m1*c*DT1 + m2*c*DT2 = 0

so you get 0.045*4190*(Tf-4) + 0.15*4190*(Tf-75) = 0

Then you just solve it algebraically.

Cehck that your answer is between 4 degrees and 75 degrees.

It is quite simple. You find m*c*DT for both of the "waters". You know the mass of both, the specific heat capacity (c) is 4190, and you know the initial temperature of both systems. You want to find the final temperature.

m1*c*DT1 + m2*c*DT2 = 0

so you get 0.045*4190*(Tf-4) + 0.15*4190*(Tf-75) = 0

Then you just solve it algebraically.

Cehck that your answer is between 4 degrees and 75 degrees.

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- #13

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~angel~ said:First of all, is it really necessary to change it to Kelvin? Is the answer meant to be in Kelvin?

It is quite simple. You find m*c*DT for both of the "waters". You know the mass of both, the specific heat capacity (c) is 4190, and you know the initial temperature of both systems. You want to find the final temperature.

m1*c*DT1 + m2*c*DT2 = 0

so you get 0.045*4190*(Tf-4) + 0.15*4190*(Tf-75) = 0

Then you just solve it algebraically.

Cehck that your answer is between 4 degrees and 75 degrees.

Yes, kelvin is necessary. Other than that this is all correct..

I just noticed I forgot the deltas in my equations above.

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