Thermodynamics - Hot Iron as a Heat Source (a) Let us consider spontaneous heat transfer from a hot reservoir 500 K into a cold reservoir 300 K, see Example 15.6 and Figure 15.34. If the total heat transfer is 9 104 J, calculate the change in entropy. (b) Consider M = 1 kg of an iron block at Tiron = 500 K and its cooling process in a room at Troom = 300 K. This heat transfer is spontaneous, from the hot iron to the cold room. Calculate the total energy provided by the hot iron in the room. We may assume that the room temperature Troom = 300 K is constant and the specific heat of iron is C = 452 J/(kgK). (Recall the stock-flow argument: the energy state of the iron changes from U500 K to U300 K. Specify the net heat transferred into the iron and the net work done by the transition.) (c) Is the above process in 2b reversible or irreversible? (d) If the heat transfer in 2b is so slow that both the initial temperature and the final temperature can be approximately equal, then the change in entropy within such an infinitesimal process is given by the following equation, see equation (15.47) in 15.6: heat into iron temperature = M C T T , (1) where we have used the fact that the heat is transferred from the iron, not into the iron, and T stands for the small temperature change. Over the whole process, the net entropy change becomes the following integral: S = M C Z Troom Tiron dT T = +M C log Tiron Troom . (2) Calculate the change in entropy. (e) If we use the hot iron block as a heat source, as we have examined in 2b, we can use U = U (500 K) U (300 K) > 0 (3) of the energy to worm the room temperature. However, we can use some energy from U to operate a heat pump and transfer more heat from the outside environment at Tenv = 273 K. Let us first estimate the maximum available work for a heat pump: U500 K W Qin / / U300 K W = U + Qin (4) 2

If the cooling is done by a reversible process, we then obtain the following maximum available work: Wmax = U Troom S, (5) where we have used the fact that the heat transfer is from the iron block to the room. In other words, Qin < 0. Calculate such the Wmax. (f) (Example 15.5) Let us estimate the best COP , wee 15.5. of a heat pump between the room Troom = 300 K and the outside environment at Tenv = 273 K. The best COP is given by the Carnot efficiency: COP = 1 Eff C , , Eff C := 1 Tenv Troom . (6) I.e., we use the outside environment as a cold heat reservoir and the room as a hot heat reservoir. Calculate COP . (g) Recalling the very definition of COP of a heat pump, see equation (15.37), we obtain Qh = COP W (7) of heat transfer from the cold outside into the room. Determine the maximum heat transfer, i.e., COP Wmax. (h) The above COP Wmax is not the total amount of heat provided by the processes above. If we write Qout = Qin (8) as the exhaust heat from the iron block to the room, we have Qout = Troom S. (9) Hence the maximum total heat available in the room is Qtotal = COP Wmax + Troom S. (10) Calculate Qtotal and compare it with the result in 2b. (i) Consider a room V = 3 m5 m8 m, about 430 sq ft with 10 feet height. The total mass of the air in this room is mair = air V (11) where air = 1.29 kg m3. Determine the change in temperature for the following cases: using the hot iron 500 K as a heat source, say 9 104 J of heat, using the ideal heat pump system above, say 3 105 J of heat. Use Cair = 900 J/(kgK).