An immersion heater of negligible mass draws a current of 7.5 A from a source with a potential difference of 120 V.The heater is in 0.950 m³ of water.The initial temperature of the water is 12.5ºC.How much time does it take to increase the temperature of the water to 27.5ºC? Neglect any heat loss.One calorie is equivalent to 4.18 J. A)4.7 $\times$ 10⁴ s B)3.3 $\times$ 10⁴ s C)1.1 $\times$ 10⁴ s D)1.6 $\times$ 10⁴ s E)None of these is correct.

Question

Quizplus · Accepted Answer

First, calculate the energy required to heat the water from 12.5ºC to 27.5ºC. The mass of water (m) is 0.950 kg (since 1 liter of water has a mass of approximately 1 kg), the specific heat capacity of water (c) is 4.18 J/gºC (or 4180 J/kgºC), and the temperature change (ΔT) is 27.5ºC - 12.5ºC = 15ºC. The energy (Q) required is Q = mcΔT = 0.950 kg * 4180 J/kgºC * 15ºC = 59,535 J.The power (P) of the heater is P = IV = 7.5 A * 120 V = 900 W (or J/s).The time (t) required is t = Q / P = 59,535 J / 900 J/s = 66.15 s.None of the provided options accurately represents this calculation.

An Immersion Heater of Negligible Mass Draws a Current of 7.5