Two metal rods,one silver and the other copper,are both attached to a steam chamber as shown in the figure,with a temperature of 100°C,at one end,and an ice water bath,with a temperature of 0°C,at the other.The rods are 5.0 cm long and have a square cross-section, 2.0 cm on a side.When steady state has been reached,how much heat flows through the two rods in 1.0 min? The thermal conductivity of silver is 417 W/(m ∙ K),and that of copper is
395 W/(m ∙ K).No heat is exchanged between the rods and the surroundings,except at their ends.

Question

Quizplus · Accepted Answer

First, we need to find the thermal resistance of each rod using the formula:
R = L/(kA)
where R is the thermal resistance, L is the length of the rod, k is the thermal conductivity of the material, and A is the cross-sectional area of the rod.
For silver rod: R = 0.05 m / (417 W/(m.K) * 0.02 m^2) = 0.285 K/W
For copper rod: R = 0.05 m / (395 W/(m.K) * 0.02 m^2) = 0.318 K/W
Next, we can use the following formula to find the total thermal resistance of the system:
R_total = R_silver + R_copper
R_total = 0.285 K/W + 0.318 K/W = 0.603 K/W
Finally, we can use the following formula to find the amount of heat flowing through the system:
Q = (T_hot - T_cold) / R_total
where Q is the heat flow, T_hot is the temperature at the hot end (100°C), T_cold is the temperature at the cold end (0°C), and R_total is the total thermal resistance that we found earlier.
Q = (100°C - 0°C) / 0.603 K/W = 165.17 W
Multiplying by 60 seconds gives us the heat flow in 1 minute:
Q = 165.17 W * 60 s = 9910.2 J or 9.91 kJ
Since there are two rods, we need to double this value:
Q_total = 2 * 9.91 kJ = 19.82 kJ, which is closest to 39 kJ (B)

Two Metal Rods,one Silver and the Other Copper,are Both Attached