Final answer:
The boiling heat transfer coefficient for the given conditions is approximately 43,500 W/m².°C, option (a), found by dividing the heat transfer rate (4.1 kW) by the product of the wire's surface area and the temperature difference (30°C).
Step-by-step explanation:
The boiling heat transfer coefficient can be determined using the rate of heat transfer equation, Q = hAΔT, where h is the heat transfer coefficient, A is the surface area, and ΔT is the temperature difference between the surface and the boiling water.
The surface area (A) of the cylindrical wire is calculated using the formula A = πDL, where D is the diameter and L is the length of the wire. Given D = 0.002 m and L = 0.5 m, the surface area is:
A = π(0.002 m)(0.5 m) = 0.00314 m².
The electric power consumption (P) is the rate of heat transfer (Q) to the water. Using the provided power P = 4.1 kW and the surface temperature of the wire at 130°C with boiling water at 100°C, the temperature difference (ΔT) is 30°C.
Calculating the heat transfer coefficient (h):
Q = hAΔT => 4.1 kW = h(0.00314 m²)(30°C)
To find h, rearrange the equation:
h = Q / (AΔT) = 4100 W / (0.00314 m² × 30°C) = 43,312.10 W/m².°C
This result most closely matches option (a) 43,500 W/m².°C. Therefore, this is the heat transfer coefficient for the given conditions, and option (a) can be considered the approximately correct answer within rounding errors.