A brass rod,75.0 cm long and having a cross-sectional area of 2.50 cm²,conducts heat from a large furnace at 375°C into a very large cold water bath at 10.0°C without losing any heat at the lateral surface of the rod.Steady state has been established,and the thermal conductivity of brass is .The rate at which the entropy of the system (furnace plus water)is changing is closest to

Question

Quizplus · Accepted Answer

The rate at which heat flows through the brass rod can be found using Fourier's Law:
q = -kA(dT/dx)

where q is the rate of heat flow, k is the thermal conductivity, A is the cross-sectional area of the rod, and (dT/dx) is the temperature gradient. Since steady state has been established and there is no heat loss at the lateral surface of the rod, the rate of heat flow through the rod is equal to the rate at which heat is transferred from the furnace to the water bath. Using the equation for the rate of entropy change in the system, we have:

dS/dt = - q/T1 + q/T2

where dS/dt is the rate of entropy change, T1 is the temperature of the furnace and T2 is the temperature of the water bath. Substituting the values given in the problem, we obtain:

dS/dt = -q/(375+273) + q/(10+273)

Solving for q, we have:

q = kA(dT/dx) = [11*(10^-3 W/(m*K))] * [(2.5*(10^-4) m^2)/(0.75 m)] * [(375-10)/(0.75 m)] = 660 W

Substituting this value of q into the equation for the rate of entropy change, we obtain:

dS/dt = -660/(375+273) + 660/(10+273) ≈ 2.64 × 10^-12 W/K

Therefore, the closest answer choice is B.

A Brass Rod,75