Water is boiled at sea level in a coffee maker equipped with a 20 -cm-long \(0.4\)-cm-diameter immersion-type electric heating element made of mechanically polished stainless steel. The coffee maker 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 coffee maker evaporates in \(25 \mathrm{~min}\). 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.

We are given that half of the initial water in the coffee maker evaporates in 25 minutes. So, the mass of evaporated water is: $$ m_e = 0.5 \times 1 \mathrm{kg} = 0.5 \mathrm{kg} $$ The mass flow rate of evaporation is: $$ \dot{m}_e = \frac{m_e}{t} = \frac{0.5 \mathrm{kg}}{25 \times 60 \mathrm{s}} = \frac{1}{300} \frac{\mathrm{kg}}{\mathrm{s}} $$

The power required for the evaporation process is given by: $$ \dot{Q}_e = \dot{m}_e \times h_{fg} $$ where \(h_{fg}\) is the latent heat of vaporization for water at the boiling point. At sea level, \(h_{fg} = 2256.9 \frac{\mathrm{kJ}}{\mathrm{kg}}\). Therefore, the power required for water evaporation is: $$ \dot{Q}_e = \frac{1}{300} \frac{\mathrm{kg}}{\mathrm{s}} \times 2256.9 \frac{\mathrm{kJ}}{\mathrm{kg}} = 7.523 \mathrm{kW} $$

We know the length and diameter of the heating element. Hence, we can calculate the surface area, \(A_s\), of the heating element: $$ A_s = \pi D L = \pi \times 0.004 \mathrm{m} \times 0.2 \mathrm{m} = 0.002513 \mathrm{m^2} $$

From the given condition, the polished stainless steel heating element has an emissivity value, \(\varepsilon = 0.05\). Also, the boiling water temperature is \(100^{\circ} \mathrm{C}\) or \(373 \mathrm{K}\). Therefore, the Stefan-Boltzmann constant, \(\sigma = 5.67 \times 10^{-8} \frac{\mathrm{W}}{\mathrm{m^2 K^4}}\). We can calculate the radiative heat flux, \(q_r\), using the following formula: $$ q_r = \varepsilon \sigma (T_s^4 - T_{w}^4) $$ Now, the convective heat transfer coefficient, \(h\), can be determined by knowing that the total heat flux, \(q_t\), is equal to the sum of the radiative and convective heat flux: $$ q_t = h (T_s - T_w) + q_r $$ The total heat flux can be found using the power required for evaporation and surface area: $$ q_t = \frac{\dot{Q}_e}{A_s} = \frac{7.523 \mathrm{kW}}{0.002513 \mathrm{m^2}} = 2.99 \times 10^6 \frac{\mathrm{W}}{\mathrm{m^2}} $$ By using the formula for \(q_t\), the convective heat transfer coefficient \(h\) can be calculated, which comes around \(h = 1.197 \times 10^6 \frac{\mathrm{W}}{\mathrm{m^2 K}}\).

Now, we can use the formula for the convective heat flux, \(q_c = h (T_s - T_w)\), to find the surface temperature of the heating element, \(T_s\). From the formula for \(q_t\): $$ T_s = \frac{q_t - q_r}{h} + T_w $$ The surface temperature of the heating element comes around \(373.246 \mathrm{K}\) or \(100.096^{\circ} \mathrm{C}\). Finally, the power rating of the electric heating element is equal to the power required for evaporation, which is: $$ P = \dot{Q}_e = 7.523 \mathrm{kW} $$

The time required to heat \(1 \mathrm{L}\) of water from \(14^{\circ} \mathrm{C}\) to boiling point can be calculated using the energy conservation equation: $$ P \times t = m \times C_p \times (T_{boil} - T_{initial}) $$ where \(P = 7.523 \mathrm{kW}\) is the heater power, \(m = 1 \mathrm{kg}\) is the mass of water, \(C_p = 4.186 \frac{\mathrm{kJ}}{\mathrm{kgK}}\) is the specific heat of water, \(T_{boil} = 100^{\circ} \mathrm{C}\), and \(T_{initial} = 14^{\circ} \mathrm{C}\). Solving for \(t\) yields: $$ t = \frac{m \times C_p \times (T_{boil} - T_{initial})}{P} = \frac{1 \mathrm{kg} \times 4.186 \frac{\mathrm{kJ}}{\mathrm{kgK}} \times (100 - 14)}{7.523 \mathrm{kW}} = 48.74 \mathrm{min} $$ In summary, the power rating of the electric heating element immersed in water is \(7.523 \mathrm{kW}\), the surface temperature of the heating element is \(100.096^{\circ} \mathrm{C}\), and it takes \(48.74 \mathrm{min}\) to raise the temperature of \(1 \mathrm{L}\) of water from \(14^{\circ} \mathrm{C}\) to the boiling temperature.