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PE Integrated Concepts (a) What is the cost of heating a

Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli ISBN: 9780130606204 3

Solution for problem 68PE Chapter 20

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Physics: Principles with Applications | 6th Edition

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Problem 68PE

PE Integrated Concepts (a) What is the cost of heating a hot tub containing 1500 kg of water from 10.0ºC to 40.0ºC , assuming 75.0% efficiency to account for heat transfer to the surroundings? The cost of electricity is 9 cents/kWh . (b) What current was used by the 220-V AC electric heater, if this took 4.00 h?

Step-by-Step Solution:

Step-by-step solution Given : Mass = 1500 kg Initial tempreature =10.0ºC Final tempreature = 40.0ºC , Efficiency = 75.0% Price= 9 cents/kWh Voltage = 220-V AC Time = 4.00 h , (1 h= 3600 seconds) (conversion) 0 Specific heat of water= 4186 J/kg C Conversion 1 kWh = 3.60 x 10 J 6 Calculate the cost of heating = Step 1 of 4 The energy required to heat water inside the tub is given by equation (1) E= m C(T - T) …………………. (1) f i Here,m is the mass of water, c is the specific heat of water,T is the final teferature and T i the initial temperature of system. Step 2 of 4 The efficiency of heating is 75%. So,compare with energy can be given by equation (2) below m C(T - f) =i5% of Q ……………(2) Rearranging equation (2) we get, equation (3) Q = mC(T - T) / (0.75) ----------------(3) f i The cost of heating water inside a hot tub is given as below equation ( 4) Cost = Q ( 12 cents / 1 kWh) ( 1 KW.h/ 3.60 x 10 J)................(4) 6 Substitute mc(T - T) / (f75) fi Q in equation (4) we get, equation (5) Cost = mcT / 0.75 (12 cents / 1kW.h) (1kWh)/ 3.60 x 10 J).............(5) 6

Step 3 of 4

Chapter 20, Problem 68PE is Solved
Step 4 of 4

Textbook: Physics: Principles with Applications
Edition: 6
Author: Douglas C. Giancoli
ISBN: 9780130606204

Physics: Principles with Applications was written by and is associated to the ISBN: 9780130606204. The full step-by-step solution to problem: 68PE from chapter: 20 was answered by , our top Physics solution expert on 03/03/17, 03:53PM. Since the solution to 68PE from 20 chapter was answered, more than 472 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Physics: Principles with Applications, edition: 6. This full solution covers the following key subjects: cost, heater, cents, concepts, containing. This expansive textbook survival guide covers 35 chapters, and 3914 solutions. The answer to “PE Integrated Concepts (a) What is the cost of heating a hot tub containing 1500 kg of water from 10.0ºC to 40.0ºC , assuming 75.0% efficiency to account for heat transfer to the surroundings? The cost of electricity is 9 cents/kWh . (b) What current was used by the 220-V AC electric heater, if this took 4.00 h?” is broken down into a number of easy to follow steps, and 58 words.

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