Part A. A room with a pine ceiling measured 12 m x 10 m x 6 cm thick. On a cold
ID: 1904251 • Letter: P
Question
Part A. A room with a pine ceiling measured 12 m x 10 m x 6 cm thick. On a cold day the temperature on inside of the ceiling is T1 = 26 oC and in the attic T2 = 7 oC. kpine = 0.120 J/msC kinsul = 0.042 J/msC. Find the rate the heat is moving through the pine ceiling. Part B. A room with a pine ceiling measured 12 m x 10 m x 6 cm thick. On a cold day the temperature on inside of the ceiling is T1 = 26 oC and in the attic T2 = 7 oC. If a 5 cm piece of insulation is added to the pine ceiling kpine = 0.120 J/msC kinsul = 0.042 J/msC. If a 5 cm piece of insulation is added to the pine ceiling , find the interface temperature for the pine and insulation. Part C. A room with a pine ceiling measured 12 m x 10 m x 6 cm thick. On a cold day the temperature on inside of the ceiling is T1 = 26 oC and in the attic T2 = 7 oC. If a 5 cm piece of insulation is added to the pine ceiling kpine = 0.120 J/msC kinsul = 0.042 J/msCIf 5 cm of insulation is added to the pine ceiling how much heat does the layer of pine / insulation transmit in one hour. Part D. A room with a pine ceiling measured 12 m x 10 m x 6 cm thick. On a cold day the temperature on inside of the ceiling is T1 = 26 oC and in the attic T2 = 7 oC. If a 5 cm piece of insulation is added to the pine ceiling kpine = 0.120 J/msC kinsul = 0.042 J/msC. If 5 cm piece of insulation is added to the pine ceiling, find the % decrease in heat loss , once the insulation has been installed.Explanation / Answer
Part A ) Rate of heat moving through the pine ceiling = K pine * A * delta T / L = 0.12 * 14*6 * 23 / 0.02 = 11592 J/s Part B : K pine * (23 - T) / 0.02 = K insu * (T - 2) / 0.06 T = 20.806 C Part C ) R = R1 + R2 = L1 / K1A1 + L2 / K2A2 = 0.019 Q = (23 - 2) / 0.019 =1105.79 J /s So heat transferred in 1 hr = 3980847.761 Joules Part D) Percentage decrease in heat loss = (11592 - 1105.79) / 11592 * 100 = 90.46 %