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I have a sealed, insulated vessel that contains water and air. I start pouring heat via an internal heater at P Watts. I know the mass/volume of air and water at the starting conditions of standard atm and temperature (1 atm, 20C). How do I calculate at what time I will reach 120C?

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    $\begingroup$ What is your attempt or thoughts on this? $\endgroup$ – user163104 Jul 24 '17 at 20:31
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In the following thread, I provided you with a cookbook recipe for determining the final equilibrium state of the system you describe, starting with specified amounts of liquid water and air in the container at 20 C, and ending with an equilibrium mixture at 120 C: Is the pressure the same if I heat to the same temperature different closed containers with distinct ratios of water to air? Applying this procedure is the first step in determining the amount of heat that has to be added.

Step 2: Evaluate the internal energy of the liquid water and the water vapor in the container at the initial and final temperatures using the steam tables. Use this to calculate the change in internal energy of the water $\Delta U_{water}$.

Step 3: Calculate the change in internal energy of the air in the container using the equation $\Delta U_{air}=mC_v\Delta T$

Step 4: Add the changes in internal energy of the water and the air together to get the total change in internal energy $\Delta U_{total}=\Delta U_{water}+\Delta U_{air}$

Step 5: Apply the first law of thermodynamics to determine the amount of heat required: $Q=\Delta U_{total}$

Step 6: Divide the heat required by the heater power P to determine the time: $t=Q/P$

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  • $\begingroup$ following this example and the one from the previous post, I arrive at an answer of 56.29 s. I am unsure if I calculated the total energy correctly (total = 84.43 kJ). I've been compiling these threads on the following google sheet if anybody is interested: docs.google.com/spreadsheets/d/… $\endgroup$ – gummibear Jul 25 '17 at 19:16
  • $\begingroup$ Looking at your tables, it seems that the energy required comes out very close to just that required to heat the initial amount of water 100 C = (200)(4.184)(100)=83680 J = 83.7 kJ $\endgroup$ – Chet Miller Jul 25 '17 at 20:00
  • $\begingroup$ Which means I am off? How exactly am I relating the internal energies for the total? I did: Specific internal energy of liquid water at 120C mass of liquid water at 120 C + Specific internal energy of gas water at 120Cmass of gas - Specific internal energy of water at 20C* original mass of water. Also, do you have a book you recommend for someone looking to learn by himself? $\endgroup$ – gummibear Jul 25 '17 at 21:14
  • $\begingroup$ No. It explains why you are right. As far as a great book is concerned, I highly recommend Fundamentals of Engineering Thermodynamics by Moran et al. I particularly like the alternate way they cover the 2nd law of thermodynamics and entropy. $\endgroup$ – Chet Miller Jul 25 '17 at 21:28
  • $\begingroup$ You have been great help Chester. Thank you and I will follow your recommendation with that book. All the best. $\endgroup$ – gummibear Jul 25 '17 at 22:19

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