# PHYSICS

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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.

• PHYSICS -

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.

• PHYSICS -

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.

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