Calculate the time in which 4.8 kJ of energy would be expended when an electric heater of resistance 1.8 x 103Ω is used on a 240 V mains supply. (Neglect he...
Question 1 Report
Calculate the time in which 4.8 kJ of energy would be expended when an electric heater of resistance 1.8 x 103Ω is used on a 240 V mains supply. (Neglect heat losses to the surrounding).
Answer Details
The power P dissipated by an electric heater of resistance R when connected to a voltage V is given by the formula: P = V2/R where P is in watts, V is in volts and R is in ohms. In this case, the electric heater has a resistance of 1.8 x 103 Ω and is connected to a 240 V mains supply. So, the power dissipated by the electric heater is: P = V2/R = (2402)/(1.8 x 103) = 32 W To calculate the time t in which 4.8 kJ of energy would be expended, we need to use the formula: E = P x t where E is the energy expended in joules, P is the power in watts, and t is the time in seconds. In this case, E = 4.8 kJ = 4,800 J and P = 32 W. Substituting these values into the formula, we get: 4,800 = 32 x t Solving for t, we get: t = 4,800 / 32 = 150 seconds Therefore, the time in which 4.8 kJ of energy would be expended when the electric heater is used on a 240 V mains supply is 150 seconds. Answer: 150.0s.
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