Question

Early Monday morning, the temperature in the lecture hall has fallen to 40° F, the same as the temperature outside. At 7:00 A.M., the janitor turns on the furnace with the thermostat set at 75 F. The time constant for the building is--2 hr and that for the building along with its heating system is hr Assuming that the outside temperature remains constant, what will be the temperature inside the lecture hall at 8:45 A.M.? When will the temperature inside the hall reach 73°F? At 8:45 AM, the temperature inside the lecture hall will be about ƯF (Round to the nearest tenth as needed.)

Answer 1

Given, T_outside = 40 degree F T_fornace = 75 degree F also, 1/k = 2 hr here, k_U = k of heating system Rightarrow k_1 + k_U = 2 Rightarrow k_U = 2 - k_1 = 2 - 1/3 Rightarrow k_U = 5/3 * we know that dT/dt = k_1(T_ouside - 7) + k_U(T_fornace - 7) Rightarrow dT/dt = 1/3(40 - T) + 5/3(75 - T) \r\n Rightarrow dT/dt = 40 - 7 + 375 - 5T/3 = 415 - 6T/3 Rightarrow dT/dt = 415/3 - 2T Rightarrow dT/dt + 2T = 415/36 Rightarrow integral dT/(138.3 - 2T) = integral dt Rightarrow ln(138.3 - 2T)/(- 2) = t + ln c Rightarrow ln(138.3 - 2T) = -2t + ln c Rightarrow ln(138.3 - 2T)/c = -2t Rightarrow 138.3 - 2T = (c) (e^- 2t) \r\n Rightarrow T = 138.3/2 - c/2 e^- 2t Rightarrow T = 69.1666 - c/2 e^- 2t here, T(0) = 40 Rightarrow 40 = 69.166 - c/2 e^- 2(0) Rightarrow c/2 = 29.166 Therefore c = 58.332 * we have to find t, at T = 73 degree F Rightarrow 78 = 69.166 - (58.332)/2 e^- 2t