A water heater consisting of a metal coil that is connected across the terminals of a 15 -V power supply is able to heat \(250 \mathrm{~mL}\) of water from room temperature to boiling point in \(45 \mathrm{~s}\). What is the resistance of the coil?

We are given the following information: - Power supply voltage (V): 15 V - Volume of water being heated (V_water): 250 mL = 0.25 L - Time taken to heat water to boiling point (t): 45 s

To find the heat energy needed to raise the water from room temperature to boiling point, we can use the formula for heat energy (Q): \(Q = mc\Delta T\) where, - Mass (m) can be calculated by multiplying the volume of water with its density, so for water, \(m = V_{water} \times \rho\), with \(\rho = 1000 \frac{kg}{m^3}\) - Specific heat capacity of water (c): 4186 J/kg·K - Temperature change (\(\Delta T\)): This is an estimation, as we do not know the exact room temperature. We can assume that the average room temperature is around 20°C and boiling point at 100°C, so the temperature change (\(\Delta T\)) is approximately 80°C. First, we calculate the mass of water: \(m = 0.25 \frac {L} {1 m^3} \times 1000 \frac {kg} {m^3}= 0.25 kg\) Now, we can calculate the heat energy required (Q): \(Q = (0.25 kg)(4186 \frac{J}{kg \cdot K})(80 K) = 83720 J\)

The heat energy we found is equal to the electrical energy supplied by the power supply. So we can use the formula for electrical power (P) to find the electrical power of the water heater: \(P = \frac{Q}{t}\) where t is the time taken to heat the water. \(P = \frac{83720 \mathrm{~J}}{45 \mathrm{~s}} = 1860 \mathrm{~W}\)

Using Ohm's law, we can find the current (I): \(I = \frac{P}{V}\) \(I = \frac{1860 \mathrm{~W}}{15 \mathrm{~V}} = 124 \mathrm{~A}\) Now applying Ohm's law once again, we can get the resistance (R): \(R = \frac{V}{I}\) \(R = \frac{15 \mathrm{~V}}{124 \mathrm{~A}} = 0.121 \Omega\) The resistance of the coil is approximately 0.121 \(\Omega\).