Mixing different temperatures of water ?

[Dmitri10]

"What would be the final temperature if you mixed a liter of 20 degrees C water with 2 liters of 40 degrees C water?"

My teacher assigned us several problems like these without bothering to explain them. I have a hunch that they are embarrassingly simple to complete, but I know not how to find the solutions. The only equation we've been given to this point is Q = mcΔT.

ANY help would be appreciated. Thanks in advance.

 

[Dick]

Q is the heat energy. The amount of heat coming out of the hotter water is equal to the amount of heat going into the cooler water. Suppose the final temperature is T (somewhere between 20 and 40 degrees). Find the change in heat for each mass of water and equate them. Then solve for T.

 

[kerimek]

I can provide you with the solution to this problem. The final temperature after mixing different temperatures of water can be calculated using the equation Q = mcΔT, where Q represents the heat transferred, m represents the mass of the water, c represents the specific heat capacity of water (which is 1 calorie/gram °C), and ΔT represents the change in temperature.

In this case, we have a total of 3 liters of water (1 liter at 20 degrees C and 2 liters at 40 degrees C), so the total mass would be 3 kg (1 kg for 1 liter of water).

To calculate the heat transferred, we can use the following formula: Q = m1c1ΔT1 + m2c2ΔT2, where m1 and ΔT1 represent the mass and change in temperature of the first water sample, and m2 and ΔT2 represent the mass and change in temperature of the second water sample.

Substituting the values, we get: Q = (1 kg)(1 cal/g °C)(20 °C) + (2 kg)(1 cal/g °C)(40 °C)
= (1 cal/g °C)(20 °C) + (2 cal/g °C)(40 °C)
= (20 cal/°C) + (80 cal/°C)
= 100 cal/°C

Now, we can use the formula Q = mcΔT to calculate the final temperature:
Q = (3 kg)(1 cal/g °C)ΔT
100 cal/°C = (3 kg)(1 cal/g °C)ΔT
ΔT = 100 cal/°C ÷ (3 kg)(1 cal/g °C)
ΔT = 33.33 °C

Therefore, the final temperature after mixing 1 liter of 20 degrees C water with 2 liters of 40 degrees C water would be 33.33 degrees C.

I hope this helps you understand the concept and how to solve similar problems in the future. If you have any other questions, please feel free to ask. Good luck with your studies!