If 42.0 kJ of heat is added to a 32.0-g sample of liquid meth-ane under 1 atm of pressure at a temperature of -170°C, what are the final state and temperature of the methane once the system equilibrates? Assume no heat is lost to the surroundings. The normal boiling point of methane is -161.5 °C. The specific heats of liquid and gaseous methane are 3.48 and 2.22 J/g-K, respectively. [Section 11.4]

Video transcript

Hello everyone today. We have the following problem. A heat amounting to 36 kg was added to a sample of liquid ammonia with a mass of g at one atmospheres and negative 40 degrees Celsius determine the final temperature and state of the ammonia. Once the system reaches equilibrium, assuming that no heat is lost to the surroundings. Used the information provided below. So we have the total heat that was added, which is 36 kg joules. And we need to calculate the heat required to raise the temperature to its boiling point. So we're gonna use the heat formula which is heat is equal to the mass times the specific capacity, times the change in temperature. Before we do that, we're gonna change all of our temperatures from Celsius to kelvin. So we have negative degrees negative 40 degrees Celsius. And to do that, we're gonna add to 73.15 kelvin. That's gonna give us two, 233.15 Kelvin. And then we're gonna take the temperature of our normal bowling point for ammonia, Which is negative 33.3°C and added to 73.15 to that to get to .85 Kelvin. So we have our equation here, Q. Eagles. The mass which we were given was 24 g. Our specific heat For this is four .74 and its jewels per gram's Kelvin. And then our change in temperature will be the two 39.85 Kelvin -233.14.15 Kelvin. And then of course we have to get rid of our units of killer jewels. So we're gonna multiply the conversion factor that one killer jewel is equal to 10 to the third jewels. When our units cancel out. We're gonna end up with the heat of 0.76 kg joules. So now we need to calculate the heat for the transition phase, transition phase from ammonia, ammonium. Any liquid to ammonium in the gasses form. So to do this we're gonna take the formula that are heat is equal to the number of moles that we have times our change and the entropy for vaporization. So to calculate this and to get moles, we're gonna take our mass or 24 g and multiply by the molar mass of ammonium. That's gonna be that one mole Is equal to 17 g according to the periodic table. We're then gonna multiply by the entropy of vaporization which is 23.4 kg joules per mole to give us a heat of 32. kg joules from there, we can calculate the excess heat that we would have. So the excess heat you would take our initial 36 kg jewels and subtract from our 0.76 kg jewels and subtract from our 32.97 kg joules to give us 2.268 kill eagles of excess heat from this. We can calculate the final temperature of our guests. So we're gonna go here and calculate the file temperature of a gas which will be our Q. Equals mass times Our specific heat times are changing temperature. We're gonna plug in our values here first though for Q. We're gonna do something left hand side here. We're going to have our 2.27 kg joules. And with that we're going to convert to regular jewels by using the conversion factor that one killer jewel is equal to 10 to the third jewels to give us 2,268.5 jules. Now that we have the heat, what we're gonna do is we're gonna plug that. So we're gonna take our We're gonna abbreviate this 2.3 times 10 to the third is equal to that 2.3 times 10 to the third, Divided by our 49. jewels per kelvin. And so from here we're gonna take this and we're gonna equal it to a value below. So we're gonna have our 49.44 joules per kelvin times our final temperature - Our initial temperature to 39.85 Kelvin. We're gonna divide by that. 49.44 And then we're gonna multiply by a 45.9 Kelvin. When we do this, we get 46 Calvin Plus of course are 2 39. equal our final temperature which is going to be 2 85.7 Kelvin or 12.6°C. So our final answer is Will be 12.6 C and ammonium as our final state. And with that we have solved the problem overall, I hope this helped, and until next time.