This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Welcome back — missed you this week. Step 1: Identify the given values and the required physical constant. The power of the heater is P = 1.08 kW. The mass of steam to be produced is m = 100 g. The water is already at 100 °C, so we only need to consider the latent heat of vaporization. The latent heat of vaporization of water (L_v) is approximately 2.26 × 10^6 J/kg. Step 2: Convert the given values to SI units. P = 1.08 kW = 1.08 × 1000 W = 1080 W m = 100 g = 100 × 10^-3 kg = 0.1 kg L_v = 2.26 × 10^6 J/kg Step 3: Calculate the total heat energy required to convert 100 g of water at 100 °C into steam. The heat energy Q required for vaporization is given by the formula Q = m L_v. Q = (0.1 kg) × (2.26 × 10^6 J/kg) Q = 2.26 × 10^5 J Step 4: Calculate the time taken for the heater to supply this amount of energy. The power P of the heater is the rate at which it supplies energy, so P = (Q)/(t), where t is the time. Rearranging for t: t = (Q)/(P) t = 2.26 × 10^5 J1080 W t = (226000)/(1080) s t ≈ 209.26 s Step 5: Compare the calculated time with the given options. The calculated time is approximately 209.26 s, which is closest to 210 s. The final answer is 210 s. Send me the next one 📸