I'm trying to figure out the logic behind working out the following exercise:

A rather lukewarm bath contains 190 litres of water at 30◦C. I add 10 litres of water at 70◦C to it in an effort to warm it up. (a) Assuming that there is no energy loss to the surroundings, what will the final temperature be?

I'm not used to solving these types of problems, but I'll try to describe my thought process:

  • The change in temperature for the more-filled bath can be shown with:

$$Q = mc_{water}\Delta T$$ $$\frac{Q}{mc_{water}} = T_f - T_i$$ $$\frac{Q}{mc_{water}} + T_i = T_f$$

  • So, I need to find $Q$, which is the heat contribution from the gas being added that hotter water, I believe. However, this leads me to some confusion: do I think along the lines of something like this?:

$$Q = mc_{water}\Delta T$$ $$Q = \rho (.2 m^3)C\Delta T$$.

I think this is wrong, since I think adding the two volumes is for the final step.

Perhaps this thinking is correct?:

$$Q_{tot} = Q_{water} + Q_{bath}$$ $$Q_{tot} = c_{water}(\rho_{water} * 0.01m^2 (70^\circ C-T_f) + \rho_{water} * 0.19m^2(30^\circ C - T_f))$$

I'm not sure if I'm justified stating this logically. Are there individual energy contributions from the temperature change respective to the water and bath that need to be taken into account? Because I think that's what this implies.

  • $\begingroup$ Just set Qtot = 0. $\endgroup$ – Chet Miller Nov 6 '17 at 2:42
  • $\begingroup$ Could you elaborate on why? $\endgroup$ – sangstar Nov 6 '17 at 2:45
  • 1
    $\begingroup$ $Q_{water}$ and $Q_{bath}$ are the changes in the heat energy of the added water and the bath water, respectively. $Q_{total}$ is the change in heat energy of all the water, which should be zero since you're just mixing the two together with the assumption that you're not adding or losing any energy while you're mixing. $\endgroup$ – user93237 Nov 6 '17 at 3:05
  • $\begingroup$ Ah, so there's a defined amount of heat energy associated with each water related to their temperature and volume. Mixing them together therefore must not add any energy to the system not from the heat energies of the waters. To imply otherwise would contradict conservation of energy - that you're getting more than what you started with. So, the change in heat energy of all the water $Q_{t0t}$, which is the summation of the heat energy of both waters, must be zero? $\endgroup$ – sangstar Nov 6 '17 at 11:30

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