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P: 789
 Quote by MoniMini 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 $$\Delta U_1=K M_1 (T_1-T)$$ and $$\Delta U_2=K M_2 (T_2-T)$$ where $\Delta U_1$ is the change in internal energy of the first case (which you don't know), $T_1$ 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 $M_1$ is the mass in the first case (1 kg). Also, $\Delta U_2$ is the change in internal energy of the second case (which you don't know), $T_2$ 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 $M_2$ 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 $$\Delta U_1+\Delta U_2=0$$ Now you have three equations and three unknowns $T$, $\Delta U_1$ and $\Delta U_2$ and you can solve for the final temperature. $$T=\frac{T_1M_1+T_2M_2}{M_1+M_2}$$ Its just a "weighted average" of the two temperatures. If both the masses were equal, you would have just the average, $(T_1+T_2)/2$