A copper sheet of thickness 2.13 mm is bonded to a steel sheet of thickness 1.41 mm. The outside surface of the copper sheet is held at a temperature of 100.0°C and the steel sheet at 24.3°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 determine the temperature of the copper-steel interface, we can use the concept of heat conduction and the principle of thermal equilibrium.
a) First, we need to calculate the heat flow through the copper and steel sheets. The heat flow can be calculated using the formula:
Q = (k * A * (T2 - T1)) / d
Where:
Q = Heat flow (in Watts)
k = Thermal conductivity of the material (in W/m°C)
A = Cross-sectional area of the sheets (in m^2)
T2 = Temperature of the outer surface (100.0°C for copper)
T1 = Temperature of the inner surface (24.3°C for steel)
d = Thickness of the sheets (in meters)
For copper:
k_copper = 401 W/m°C
A_copper = 1 m^2
T2_copper = 100.0°C
T1_copper = T (unknown temperature at the interface)
d_copper = 2.13 mm = 0.00213 m
For steel:
k_steel = 50 W/m°C
A_steel = 1 m^2
T2_steel = T (unknown temperature at the interface)
T1_steel = 24.3°C
d_steel = 1.41 mm = 0.00141 m
Now we can equate the heat flow for both copper and steel to find the temperature at the interface:
Q_copper = Q_steel
(k_copper * A_copper * (T2_copper - T1_copper)) / d_copper = (k_steel * A_steel * (T2_steel - T1_steel)) / d_steel
Substituting the given values and solving for T:
(401 * 1 * (100.0 - T)) / 0.00213 = (50 * 1 * (T - 24.3)) / 0.00141
Simplifying the equation gives:
2.13(100 - T) = 0.141(T - 24.3)
213 - 2.13T = 0.141T - 3.43
2.13T + 0.141T = 213 + 3.43
2.271T = 216.43
T = 95.46°C
Therefore, the temperature at the copper-steel interface is approximately 95.46°C.
b) To calculate the heat conducted through 1.00 m^2 of the combined sheets per second, we can use the formula:
Q = (k_eff * A * (T2 - T1)) / d
Where:
Q = Heat flow (in Watts)
k_eff = Effective thermal conductivity, obtained by using the formula:
1/k_eff = (d1/k1) + (d2/k2)
Here, d1 and d2 are the thicknesses of copper and steel respectively, and k1 and k2 are the thermal conductivities of copper and steel respectively.
d1 = 2.13 mm = 0.00213 m
d2 = 1.41 mm = 0.00141 m
k1 = 401 W/m°C (for copper)
k2 = 50 W/m°C (for steel)
Substituting the values into the formula for k_eff:
1 / k_eff = (0.00213 / 401) + (0.00141 / 50)
k_eff = 1 / ((0.00213 / 401) + (0.00141 / 50))
Now we can calculate the heat flow (Q) using the effective thermal conductivity:
Q = (k_eff * A * (T2 - T1)) / d
Q = (k_eff * 1 * (100.0 - 24.3)) / (0.00213 + 0.00141)
Substituting the values and calculating, we can find the heat conducted through 1.00 m^2 of the combined sheets per second.
Please note that the specific numerical value cannot be determined without knowing the value of k_eff and the dimensions of A.