Question

Water is boiled at sea level in a coffeemaker equipped with a \(30-\mathrm{cm}\)-long, 0.4-cm-diameter immersion-type electric heating element made of mechanically polished stainless steel. The coffeemaker initially contains \(1 \mathrm{~L}\) of water at \(14^{\circ} \mathrm{C}\). Once boiling starts, it is observed that half of the water in the coffeemaker evaporates in \(32 \mathrm{~min}\). Determine the power rating of the electric heating element immersed in water and the surface temperature of the heating element. Also determine how long it will take for this heater to raise the temperature of $1 \mathrm{~L}\( of cold water from \)14^{\circ} \mathrm{C}$ to the boiling temperature.

Short answer

2. What is the approximate surface temperature of the heating element? 3. How long does it take for the heater to raise the temperature of 1 L of water from 14°C to boiling temperature?

Step-by-step solution

Calculate the heat required to raise the temperature of half a liter of water to boiling temperature.

To raise the temperature of half a liter (500 g) of water from 14°C to boiling temperature, we need to determine the amount of heat, Q, required using the formula Q = mcΔT: - m: mass of water (0.5 kg) - c: specific heat capacity of water (approximately 4200 J/(kg·K)) - ΔT: change in temperature (100 - 14 = 86°C) Q = (0.5 kg) × (4200 J/(kg·K)) × (86 K) ≈ 180600 J

Calculate the rate of heat transfer for the evaporation process.

To evaporate half a liter of water in 32 minutes, we need to find the rate of heat transfer, P, during the evaporation process: - L: latent heat of vaporization of water (approximately 2260 kJ/kg) - t: time taken for the evaporation process (32 * 60 = 1920 s) P = (0.5 kg × 2260 kJ/kg × 1000 J/kJ + 180600 J) / 1920 s ≈ 586 W

Determine the power rating of the electric heating element.

Since 586 W is the power required for the evaporation of water, we can assume that this is also the power rating of the electric heating element. Thus, the power rating of the electric heating element is P = 586 W.

Calculate the surface temperature of the heating element.

To find the surface temperature of the heating element, we can use the formula for steady-state conduction through a cylinder: P = (2 * pi * L * k * (T_surface - T_water)) / ln(r_outer/r_inner) Where: - L = length of heating element (30 cm = 0.3 m) - k = thermal conductivity of stainless steel (approximately 18 W/(m·K)) - T_surface = surface temperature of heating element - T_water = temperature of water at boiling point (100°C) - r_outer = outer radius of the heating element (0.2 cm = 0.002 m) - r_inner = inner radius of heating element (applicable only for hollow cylinders, but in our case, assume r_inner to be negligible) We can solve this for T_surface: T_surface = (P * ln(r_outer/r_inner) + T_water * 2 * pi * L * k) / (2 * pi * L * k) + T_water Assuming r_inner to be negligible, the ln(r_outer/r_inner) term becomes ln(1/r_outer) = ln(1/0.002) ≈ 6.21. We can substitute our values: T_surface = (586 W * 6.21 + 100°C * 2 * pi * 0.3 m * 18 W/(m·K)) / (2 * pi * 0.3 m * 18 W/(m·K)) ≈ 152.66°C

Calculate the time required to raise the temperature of 1 L of water to boiling.

Finally, we can find the time required to raise the temperature of 1 L of water from 14°C to boiling temperature by first calculating the heat required, and then dividing by the power of the heating element: Q2 = (1 kg) × (4200 J/(kg·K)) × (86 K) ≈ 361200 J t2 = Q2 / P = 361200 J / 586 W ≈ 616 s So, it will take approximately 616 seconds (or about 10 minutes and 16 seconds) for the heater to raise the temperature of 1 L of water from 14°C to boiling temperature. To summarize: - The power rating of the electric heating element is 586 W. - The surface temperature of the heating element is approximately 152.66°C. - It will take about 10 minutes and 16 seconds for the heating element to raise the temperature of 1 L of water from 14°C to boiling temperature.