Chapter 2.

Note that at 1500°C heat can be added to take the iron between points H and F without changing its temperature at all so this is a phase transition (bond energy changes but not thermal energy). Because it is the lowest temperature phase transition it must be the melting transition. The situation above describes iron at a temperature well below the melting temperature and the only point here is point I where the iron is still in the solid state (T < Tmelt).

You probably know that water melts at 0°C and boils at 100°C. The temperature of the water is on the vertical axis and we find T = 100°C for each of points B, C, and D. Point D represents liquid water at 100°C, point B represents gaseous water (steam) at 100°C, and point C represents a mixture of liquid and gas. At point C about half the heat required to fully vaporize the water has been added so about half of the liquid has been turned into gas and that is the situation specified.

The vertical axis has the relevant temperatures. The regions of the graph where you can add heat without changing the temperature (the horizontal lines) are the phase-change temperatures. Pure iron (in equilibrium at standard pressure) has only one possible melting temperature and only one possible boiling temperature so diagram C which has a melting temperature a little above 1500°C and a boiling temperature around 2800°C matches those temperatures for iron shown in diagram A so C is also for iron. The heat added (horizontal line) is different because that depends on how much iron we have (about 1.5 times as much iron gives the heats shown in diagram C). Diagram B is for a piece of copper, which has lower melting and boiling temperatures. We made the numbers on the horizontal axis approximately the same as diagram A by choosing just the right amount of copper.

a) Reading on the y-axis of diagram A shows us that Ti iron = 750° C is at a temperature close to the point labeled I. The iron is a solid at this temperature. b) Reading on the y-axis of diagram B shows us that Ti water = -50° C is very close to point I. The water is a solid (ice) at this temperature. c) The iron is initially hotter (higher temperature) than the water so it will lose heat energy to the lower T ice. The x-axis is a measure of heat added with more heat added being to the right so if heat is taken out the iron must move to the left (opposite of right) and down (i.e. iron cools off). d) The water is initially cooler (lower temperature) than the iron so it will gain heat energy from the higher T iron. The x-axis is a measure of heat added with more heat added being to the right so the water must move to the right and up. e) According to the graph, 20 J of heat transferred away from the iron moves the iron down to somewhere around 0°C. Since the two materials are supposed to be at “equilibrium”, they should have the same temperature so if the iron is at T = 0°C then the water must be at T = 0°C. Is this possible? Yes, since the iron lost 20 J of energy then the water must have gained 20 J (because energy is conserved) and, according to the graph for water, this addition of 20 J of energy would leave the water at about 0°C and probably mostly still solid. It is hard to read the numbers very precisely on these graphs so i) there might be a little melting with an equilibrium T = 0°C OR ii) the final temperature might be a little less than 0°C in which case the water would all still be solid.