A copper sheet of thickness 2.19 mm is bonded to a steel sheet of thickness 1.13 mm. The outside surface of the copper sheet is held at a temperature of 100.0°C and the steel sheet at 25.7°C.
a) Determine the temperature (in °C) of the copper-steel interface.
b) How much heat is conducted through 1.00 m2 of the combined sheets per second?
To solve this problem, we need to use the formula for heat conduction through a composite material. The formula is:
q = (k * A * (T1 - T2)) / d
Where:
q is the heat conducted per unit of time (in this case, per second)
k is the thermal conductivity of the material
A is the surface area through which the heat is conducted
T1 and T2 are the temperatures of the two surfaces
d is the total thickness of the composite material
First, let's use the given information to calculate the temperature of the copper-steel interface (T_interface).
a) Determine the temperature (in °C) of the copper-steel interface.
1. Convert the temperatures to Kelvin by adding 273.15:
T_copper = 100.0 + 273.15 = 373.15 K
T_steel = 25.7 + 273.15 = 298.85 K
2. Calculate the temperature of the interface using the formula:
T_interface = (T_copper + T_steel) / 2
T_interface = (373.15 + 298.85) / 2 = 336 K
Therefore, the temperature of the copper-steel interface is 336°C.
b) How much heat is conducted through 1.00 m2 of the combined sheets per second?
1. Calculate the total thickness of the composite material:
d = thickness of copper + thickness of steel
d = 2.19 mm + 1.13 mm = 3.32 mm = 0.00332 m
2. Look up the thermal conductivity values for copper (kcopper) and steel (ksteel). Let's assume that kcopper = 401 W/(m·K) and ksteel = 50.2 W/(m·K).
3. Calculate the heat conducted per second using the formula:
q = (k * A * (T1 - T2)) / d
For simplicity, let's assume A = 1.00 m^2.
q = [kcopper * A * (T_copper - T_interface) / thickness of copper] + [ksteel * A * (T_interface - T_steel) / thickness of steel]
q = [(401 * 1.00 * (373.15 - 336)) / 0.00219] + [(50.2 * 1.00 * (336 - 298.85)) / 0.00113]
Simplify the equation:
q = (105,051.14 / 0.00219) + (32,714.14 / 0.00113)
q = 47,948,175.34 + 28,004,560.18
q = 75,952,735.52 W
Therefore, the amount of heat conducted through 1.00 m^2 of the combined sheets per second is approximately 75,952,735.52 Watts or Joules per second.