Two metal rods -- one copper the other aluminum -- are connected in series, as shown in the figure below. Note that each rod is 0.525 m in length, and has a square cross section 1.25 cm on a side. The temperature at the copper end of the rods is 2.00°C; the temperature at the aluminum end is 106°C.
The average temperature of the two ends is 54.0°C.
(b) Given that the heat flow through each of these rods in 1.00 s is 4.34 J, find the temperature at the copper-aluminum interface.
I assume they are talking about a steady state heat flow situation. The far end of the copper rod is 2 C and the far end of the Al rod is 106 C. It is obvious that the average temperature of the two far ends is (106+2)/2 = 54 C. That is not independent information.
In steady state, require that the heat flow in both rods be the same 4.34 Watts and solve for the temperature T of the interface (where they touch).
(T-2)*Kcu*A/L = (106-T)*Kal*A/L = 4.34 Watts
A is the cross sectional area and L is the length. Look up the thermal conductivities and solve for T.
You can cancel out A/L and solve for T directly, without having to know the heat flow.
(T-2)*Kcu= (106-T)*Kal
T-2 = (Kal/Kcu)*(106-T) = r(106-T)
where r= Kal/Kcu
T(1 + r) = 2 + r*106
T = (2 + 106r)/(1+r)
They have provided you a lot of unecessary information to reach a solution (A, L and heat flow), but you will have to look up the thermal conductivities that are required.
The heat flow value that they provided may actually be wrong. It will determined by what the k values are.
Five-gram samples of copper and aluminum are at room temperature. Both receive equal amounts of energy due to heat flow. The specific heat capacity of copper is 0.09 cal/g°C, and the specific heat capacity of aluminum is 0.22 cal/g°C. Which of the following statements is true?
The temperature of each sample increases by the same amount.
The aluminum will get hotter than the copper.
The copper will get hotter than the aluminum.
The temperature of each sample decreases by the same amount.
To find the temperature at the copper-aluminum interface, we can use the principle of thermal equilibrium. In thermal equilibrium, the heat flow through each rod will be equal.
Let's assume that the heat flow through the copper rod is Qc, and the heat flow through the aluminum rod is Qa.
Given that the heat flow through each rod is 4.34 J in 1.00 s, we can write:
Qc = 4.34 J
Qa = 4.34 J
Since the heat flow through each rod is equal, we can write:
Qc = Qa
Now, let's consider the rate of heat flow using the formula:
Q = k * A * (T2 - T1) / L
Where:
Q is the heat flow
k is the thermal conductivity
A is the cross-sectional area of the rod
T2 is the final temperature of the rod
T1 is the initial temperature of the rod
L is the length of the rod
For the copper rod:
Qc = kCopper * A * (Tcopper_interface - 2°C) / 0.525 m (Equation 1)
For the aluminum rod:
Qa = kAluminum * A * (106°C - Taluminum_interface) / 0.525 m (Equation 2)
Since Qc = Qa, we can equate both equations:
kCopper * A * (Tcopper_interface - 2°C) / 0.525 m = kAluminum * A * (106°C - Taluminum_interface) / 0.525 m
A and L cancel out, so we get:
kCopper * (Tcopper_interface - 2°C) = kAluminum * (106°C - Taluminum_interface)
Rearranging the equation, we have:
kCopper * Tcopper_interface - 2 * kCopper = 106 * kAluminum - kAluminum * Taluminum_interface
Combining like terms, we get:
kCopper * Tcopper_interface + kAluminum * Taluminum_interface = 106 * kAluminum + 2 * kCopper
Since average temperature (54.0°C) = (Tcopper_interface + Taluminum_interface) / 2, we can substitute Tcopper_interface = 54.0°C - Taluminum_interface into the equation:
kCopper * (54.0°C - Taluminum_interface) + kAluminum * Taluminum_interface = 106 * kAluminum + 2 * kCopper
Now, we have an equation with a single unknown (Taluminum_interface). We can solve for Taluminum_interface.
0.525 * kCopper = 106 * kAluminum + 2 * kCopper
Simplifying the equation further:
0.525 * kCopper - 2 * kCopper = 106 * kAluminum
(0.525 - 2) * kCopper = 106 * kAluminum
-1.475 * kCopper = 106 * kAluminum
Taluminum_interface = (106 * kAluminum) / (1.475 * kCopper)
To find the value of Taluminum_interface, we need to know the thermal conductivities of copper (kCopper) and aluminum (kAluminum). If you have these values, substitute them into the equation to find the temperature at the copper-aluminum interface.
To find the temperature at the copper-aluminum interface, we will use the concept of thermal conductivity and the equation for heat flow.
The equation for heat flow through a rod is:
Q = k * A * (ΔT / L) * t
Where:
- Q is the heat flow
- k is the thermal conductivity
- A is the cross-sectional area of the rod
- ΔT is the temperature difference across the rod
- L is the length of the rod
- t is the time
We are given that the heat flow through each of the rods in 1.00 s is 4.34 J. Let's assume that the heat flow is the same through both rods.
Now, we can rearrange the equation to solve for the temperature difference (ΔT) across the copper-aluminum interface:
ΔT = Q * L / (k * A * t)
Since we know the values for Q (4.34 J), L (0.525 m), A (1.25 cm x 1.25 cm), t (1.00 s), we need to find the thermal conductivities (k) of copper and aluminum to calculate ΔT.
The thermal conductivity of copper is approximately 401 W/(m·K) and the thermal conductivity of aluminum is approximately 205 W/(m·K).
Now let's plug in the values into the equation:
ΔT = (4.34 J) * (0.525 m) / [(401 W/(m·K) + 205 W/(m·K)) * (0.0125 m * 0.0125 m) * (1.00 s)]
Simplifying the equation and calculating:
ΔT ≈ 0.0088 K
Finally, to find the temperature at the copper-aluminum interface, we can subtract this temperature difference from the temperature at the copper end:
Temperature at the copper-aluminum interface = 2.00°C - 0.0088 K
Therefore, the temperature at the copper-aluminum interface is approximately 1.99°C.