Mixing water at different temperatures

In summary, assuming that no heat is lost to the surrroundings, the final temperature when 1 kg of water at 10°C is mixed with 5 kg of water at 80°C will be 63.8°C.
  • #1
MoniMini
12
0
Hello all,
I'm having trouble solving the given problem.

"Assuming that no heat is lost to the surrroundings, what will be the final temperature when 1 kg of water at 10°C is mixed with 5 kg of water at 80°C"

Taking the mean of the 2 fluids will not help here since their mass is different.
I don't know any other formula regarding such problems.

Thank You,
~MoniMini
 
Science news on Phys.org
  • #2
MoniMini said:
Hello all,
I'm having trouble solving the given problem.

"Assuming that no heat is lost to the surrroundings, what will be the final temperature when 1 kg of water at 10°C is mixed with 5 kg of water at 80°C"

Taking the mean of the 2 fluids will not help here since their mass is different.
I don't know any other formula regarding such problems.

Thank You,
~MoniMini

You can say that the specific heat of water is constant. This is not exactly right, but it will be very close. That means the change in internal energy is proportional to its change in temperature. Its also proportional to the amount of water you have. So [tex]\Delta U_1=K M_1 (T_1-T) [/tex] and [tex]\Delta U_2=K M_2 (T_2-T) [/tex] where [itex]\Delta U_1[/itex] is the change in internal energy of the first case (which you don't know), [itex]T_1[/itex] is the initial temperature for the first case (10°C) , T is the final temperature for the first case (what you are trying to find), and [itex]M_1[/itex] is the mass in the first case (1 kg). Also, [itex]\Delta U_2[/itex] is the change in internal energy of the second case (which you don't know), [itex]T_2[/itex] is the initial temperature for the second case (80°C) , T is the final temperature for the second case (what you are trying to find), and [itex]M_2[/itex] is the mass in the second case (5 kg). K is some constant, but don't worry about it, it cancels out. Notice that the final temperature is T in both cases, because when you mix them together, they both go to the same temperature. Finally, you know that no heat was added or subtracted, so the total change in internal energy has to be zero. That means [tex]\Delta U_1+\Delta U_2=0[/tex] Now you have three equations and three unknowns [itex]T[/itex], [itex]\Delta U_1[/itex] and [itex]\Delta U_2[/itex] and you can solve for the final temperature. [tex]T=\frac{T_1M_1+T_2M_2}{M_1+M_2}[/tex] Its just a "weighted average" of the two temperatures. If both the masses were equal, you would have just the average, [itex](T_1+T_2)/2[/itex]
 
Last edited:
  • #3
I found the specific heat capacity online. http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html

The heat capacity barely changes from 10 to 80 degrees.

But if you wanted to factor the heat capacity change, you could integrate the heat capacity across temperature to get internal energy versus temperature per mass. Then add up the initial energies and this is equal to your final energy.
 
  • #4
Rap said:
You can say that the specific heat of water is constant. This is not exactly right, but it will be very close. That means the change in internal energy is proportional to its change in temperature. Its also proportional to the amount of water you have. So [tex]\Delta U_1=K M_1 (T_1-T) [/tex] and [tex]\Delta U_2=K M_2 (T_2-T) [/tex] where [itex]\Delta U_1[/itex] is the change in internal energy of the first case (which you don't know), [itex]T_1[/itex] is the initial temperature for the first case (10°C) , T is the final temperature for the first case (what you are trying to find), and [itex]M_1[/itex] is the mass in the first case (1 kg). Also, [itex]\Delta U_2[/itex] is the change in internal energy of the second case (which you don't know), [itex]T_2[/itex] is the initial temperature for the second case (80°C) , T is the final temperature for the second case (what you are trying to find), and [itex]M_2[/itex] is the mass in the second case (5 kg). K is some constant, but don't worry about it, it cancels out. Notice that the final temperature is T in both cases, because when you mix them together, they both go to the same temperature. Finally, you know that no heat was added or subtracted, so the total change in internal energy has to be zero. That means [tex]\Delta U_1+\Delta U_2=0[/tex] Now you have three equations and three unknowns [itex]T[/itex], [itex]\Delta U_1[/itex] and [itex]\Delta U_2[/itex] and you can solve for the final temperature. [tex]T=\frac{T_1M_1+T_2M_2}{M_1+M_2}[/tex] Its just a "weighted average" of the two temperatures. If both the masses were equal, you would have just the average, [itex](T_1+T_2)/2[/itex]

Thanks a TON! You explained it really well.
I got the correct anser. I don't know how to type in the mathematical symbols and all, but on paper I calculated and the result was 63.8°C which is the correct answer.
Thanks again :)

~MoniMini
 
  • #5


Hello MoniMini,

I can offer some insights into this problem. When mixing two substances at different temperatures, the final temperature will depend on the mass and initial temperatures of each substance. In this case, we have 1 kg of water at 10°C and 5 kg of water at 80°C.

To find the final temperature, we can use the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred. In this case, the total energy of the system (the combined water) will remain constant. This means that the heat lost by the hot water (80°C) will be equal to the heat gained by the cold water (10°C).

We can use the formula Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Since we are assuming no heat loss to the surroundings, we can set Q = 0.

Solving for the final temperature, we get:

(5 kg)(4.186 J/g°C)(80°C - T) = (1 kg)(4.186 J/g°C)(T - 10°C)

Simplifying, we get:

20.93 kg°C - 4.186 kgT = 4.186 kgT - 41.86 kg°C

9.372 kgT = 62.79 kg°C

T = 6.7°C

Therefore, the final temperature when 1 kg of water at 10°C is mixed with 5 kg of water at 80°C is 6.7°C. I hope this helps you understand the problem better. Let me know if you have any further questions. Good luck!
 

1. How does mixing water at different temperatures affect the overall temperature?

When mixing water at different temperatures, the resulting temperature will depend on the quantities and temperatures of the individual waters. The final temperature will be somewhere between the two initial temperatures, with the warmer water losing heat and the cooler water gaining heat until they reach a thermal equilibrium.

2. Is it safe to mix hot and cold water when preparing a beverage or food?

Yes, it is generally safe to mix hot and cold water for food or beverages. However, it is important to make sure the hot water is not boiling and the cold water is not too cold to avoid extreme temperatures.

3. Does the type of container used for mixing water at different temperatures affect the overall temperature?

Yes, the type of container used can affect the overall temperature. For example, a metal container will conduct heat more quickly than a plastic container, resulting in a faster temperature change.

4. Can mixing water at different temperatures change its chemical properties?

No, mixing water at different temperatures will not alter its chemical properties. The only change will be in the temperature and possibly the physical properties, such as density.

5. What is the best way to mix water at different temperatures to achieve a desired temperature?

The best way to mix water at different temperatures to achieve a desired temperature is to use a thermometer and add small amounts of each water while continuously checking the temperature until the desired temperature is reached. This will ensure that the final temperature is as accurate as possible.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
967
  • Thermodynamics
Replies
4
Views
880
Replies
10
Views
5K
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
17
Views
2K
Replies
25
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
906
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
1K
Replies
3
Views
1K
Back
Top