Question

A copper sheet of thickness \(2.00 \mathrm{~mm}\) is bonded to a steel sheet of thickness \(1.00 \mathrm{~mm}\). The outside surface of the copper sheet is held at a temperature of \(100.0^{\circ} \mathrm{C}\) and the steel sheet at \(25.0^{\circ} \mathrm{C} .\) a) Determine the temperature of the copper-steel interface. b) How much heat is conducted through \(1.00 \mathrm{~m}^{2}\) of the combined sheets per second?

Short answer

Answer: The temperature of the copper-steel interface is approximately 88.4°C, and the heat conducted through 1.00 m² of combined sheets per second is about 14,580 W or 14.58 kW.

Step-by-step solution

Identify given values

We are given the following values: Thickness of the copper sheet \(d_C = 2.00 \mathrm{~mm}\) Thickness of the steel sheet \(d_S = 1.00 \mathrm{~mm}\) Temperature of copper sheet's outer surface \(T_C = 100.0^{\circ} \mathrm{C}\) Temperature of steel sheet's outer surface \(T_S = 25.0^{\circ} \mathrm{C}\) We need to find the temperature of the copper-steel interface and the heat conducted through a \(1.00 \mathrm{~m}^{2}\) combined sheets per second.

Convert given values to proper units

Convert the values of thickness into meters. \(d_C = 2.00 \times 10^{-3} \mathrm{~m}\) \(d_S = 1.00 \times 10^{-3} \mathrm{~m}\)

Apply Fourier's Law of conduction

In steady-state conditions, the heat flow (heat per second) through the copper sheet must be equal to the heat flow through the steel sheet. We can write Fourier's Law of conduction as: \(Q = kA \frac{(T_{hot} - T_{cold})}{d}\) where \(Q\) is the heat flow (W), \(k\) is the thermal conductivity (W/mK), \(A\) is the cross-sectional area (m²), \(T_{hot}\) and \(T_{cold}\) are the temperatures at the two ends of the sheet (K), \(d\) is the thickness (m). We have two equations for heat flow through copper and steel, respectively: \(Q_C = k_CA\frac{(T_C - T_{i})}{d_C}\) \(Q_S = k_SA\frac{(T_{i} - T_S)}{d_S}\) where \(Q_C = Q_S\), \(T_{i}\) is the temperature of the copper-steel interface, \(k_C\) and \(k_S\) are the thermal conductivities of copper and steel, respectively.

Calculate the temperature of the copper-steel interface

Since \(Q_C = Q_S\), we can equate the two equations from Step 3: \(k_CA \frac{(T_C - T_{i})}{d_C} = k_SA \frac{(T_{i} - T_S)}{d_S}\) Rearrange the equation to solve for \(T_{i}\): \(T_{i} = \frac{d_Sk_CT_C + d_Ck_ST_S}{d_Ck_S + d_Sk_C}\) In order to use this equation, we need to know the values of the copper and steel thermal conductivities. The thermal conductivities are: Copper: \(k_C = 398 \mathrm{~W/mK}\) Steel: \(k_S = 50 \mathrm{~W/mK}\) Now, we can plug in the known values into the equation and solve for \(T_{i}\): \(T_{i} = \frac{(1.00 \times 10^{-3} \mathrm{~m}) (398 \mathrm{~W/mK}) (100.0^{\circ} \mathrm{C}) + (2.00 \times 10^{-3} \mathrm{~m}) (50 \mathrm{~W/mK}) (25.0^{\circ} \mathrm{C})}{(2.00 \times 10^{-3} \mathrm{~m})(50 \mathrm{~W/mK}) + (1.00 \times 10^{-3} \mathrm{~m})(398 \mathrm{~W/mK})}\) \(T_{i} \approx 88.4^{\circ} \mathrm{C}\) The temperature of the copper-steel interface is \(88.4^{\circ} \mathrm{C}\).

Calculate the heat conducted

Now, we can find the heat conducted through \(1.00 \mathrm{~m}^{2}\) of the combined sheets per second. We can use the heat flow equation through either the copper or steel sheet. Let's use the copper sheet for calculation: \(Q = k_CA\frac{(T_C - T_{i})}{d_C}\) Replace the known values into the equation: \(Q = (398 \mathrm{~W/mK})(1.00 \mathrm{~m}^2) \frac{(100.0^{\circ} \mathrm{C} - 88.4^{\circ} \mathrm{C})}{2.00 \times 10^{-3}\mathrm{~m}}\) \(Q \approx 14,580 \mathrm{~W}\) The heat conducted through \(1.00 \mathrm{~m}^{2}\) of the combined sheets per second is approximately \(14,580 \mathrm{~W}\) or \(14.58 \mathrm{~kW}\).

Key concepts

Heat Transfer

Heat transfer is a fundamental concept in physics that deals with the movement of thermal energy from one place to another. It occurs in three main ways: conduction, convection, and radiation. In the context of our exercise, the focus is on conduction, which is the transfer of heat through a solid material without any actual movement of the substance.

In the example given, heat is conducted through a copper sheet bonded to a steel sheet, and understanding how heat is transferred in such scenarios is crucial for various engineering applications. Heat flows from the hot surface of the copper at 100.0°C towards the cooler steel sheet at 25.0°C until thermal equilibrium is reached.

Thermal Conductivity

Thermal conductivity, represented by the symbol \(k\), is a material property that measures the ability to conduct heat. It is defined as the amount of heat that passes in unit time through unit area of a substance with a temperature gradient of one degree per unit distance. High \(k\) values indicate good conductors of heat, such as metals like copper and aluminum, while low \(k\) values denote poor conductors or insulators, like rubber or glass.

In the exercise we've tackled, copper has a \(k\) of 398 W/mK, which is significantly higher than steel's \(k\) of 50 W/mK, explaining why copper is often used in thermal applications, where efficient heat transfer is necessary.

Steady-State Conduction

Steady-state conduction refers to a condition where the temperature field within a conducting medium remains constant over time. This means that any point within the medium has a temperature that does not change, indicating that the heat entering a particular section is equal to the heat leaving it.

In our exercise, we assume steady-state conditions to simplify the calculations, allowing us to use Fourier's Law of conduction to find the temperature at the copper-steel interface and the heat flow through the sheets. We calculated the interface temperature as 88.4°C and the heat flow as approximately 14,580 W or 14.58 kW per square meter. Understanding steady-state conduction is vital in designing systems involving heat exchange, such as heat exchangers, insulation, and electronic devices.