A copper sheet of thickness 2.01 mm is bonded to a steel sheet of thickness 1.07 mm. The outside surface of the copper sheet is held at a temperature of 100.0°C and the steel sheet at 22.9°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 secon

WEll, heat flows from the outside of the steel to the outside of the copper.

You need the thermal conductivity coefficents for steel, and copper. I will use Cs, and Cc for those coefficents.

Assume Ti for the temperature of the interface.

And finally, you know the heat moving across the steel is the same as heat moving across the copper.

HeatacrossSteel=HeatacrossCopper
(100-Ti)Cs*.00107=(Ti-22.9)Cc*.00201

solve for Ti check my work.

http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html

A copper sheet of thickness 2.49 mm is bonded to a steel sheet of thickness 1.51 mm. The outside surface of the copper sheet is held at a temperature of 100.0°C and the steel sheet at 24.1°C.

To determine the temperature of the copper-steel interface, we can use the concept of thermal conductivity. Thermal conductivity is a property of a material that describes its ability to conduct heat.

To find the temperature at the interface, we first need to calculate the heat flux across the copper and steel sheets. The heat flux can be determined using Fourier's Law of Heat Conduction, which states that the heat flux (Q) through a material is proportional to the temperature difference (ΔT) across the material and inversely proportional to the material's thermal resistance (R).

Q = (ΔT / R)

To calculate the heat flux through the combined sheets, we can use the thermal resistance formula:

R = (d / k)

Where:
- R is the thermal resistance
- d is the thickness of the material
- k is the thermal conductivity of the material

For the copper sheet:
- thickness (d) = 2.01 mm = 0.00201 m
- thermal conductivity (k_copper) = 386 W/(m·K) (at room temperature)

For the steel sheet:
- thickness (d) = 1.07 mm = 0.00107 m
- thermal conductivity (k_steel) = 43 W/(m·K) (at room temperature)

Since the heat flux through both materials will be the same (assuming a steady-state condition), we can set up the following equation:

(ΔT_copper / R_copper) = (ΔT_steel / R_steel)

Substituting the values, we get:

(ΔT_copper / (d_copper / k_copper)) = (ΔT_steel / (d_steel / k_steel))

Simplifying, we find:

ΔT_copper / (d_copper / k_copper) = ΔT_steel / (d_steel / k_steel)

Now, we know the temperature on each side of the copper and steel sheets:

- Temperature of copper side: T_copper = 100.0°C
- Temperature of steel side: T_steel = 22.9°C

Substituting these values into the equation, we have:

(T_copper - T_interface) / (d_copper / k_copper) = (T_interface - T_steel) / (d_steel / k_steel)

We can rearrange the equation and solve for T_interface, which is the temperature at the copper-steel interface:

T_interface = ((T_copper / (d_copper / k_copper)) + (T_steel / (d_steel / k_steel))) / ((1 / (d_copper / k_copper)) + (1 / (d_steel / k_steel)))

Now, let's plug in the values:

T_interface = ((100.0°C / (0.00201 m / 386 W/(m·K))) + (22.9°C / (0.00107 m / 43 W/(m·K)))) / ((1 / (0.00201 m / 386 W/(m·K))) + (1 / (0.00107 m / 43 W/(m·K))))

After performing the calculations, we can find the temperature at the copper-steel interface.

To determine the amount of heat conducted through 1.00 m2 of the combined sheets per second, we can use the formula:

Q = k * A * ΔT / d

Where:
- Q is the heat transferred per unit time (in this case, per second)
- k is the thermal conductivity of the combined sheets
- A is the area of the sheets
- ΔT is the temperature difference across the sheets
- d is the combined thickness of the copper and steel sheets

Since the heat flux through both materials will be the same (assuming a steady-state condition), the total heat flux passing through the combined sheets can be calculated as follows:

Q = (k_copper * A * ΔT) / d_copper
+ (k_steel * A * ΔT) / d_steel

Substituting the given values:

Q = (386 W/(m·K) * 1.00 m^2 * (100.0°C - T_interface)) / 0.00201 m
+ (43 W/(m·K) * 1.00 m^2 * (T_interface - 22.9°C)) / 0.00107 m

After performing the calculations, we can determine the amount of heat conducted through 1.00 m2 of the combined sheets per second.