Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) What is the cost of heating a hot tub containing 1500kg of water from to assuming 75.0% efficiency to account for heat transfer to the surroundings? The cost of electricity is 12cents/centskWโ‹…hkWโ‹…h.(b) What current was used by the 220โˆ’VAC electric heater, if this took 4.00h?

Short Answer

Expert verified
  1. Cost of the energy supplied to heater is$6.28.
  2. Amount of current used by heater is 79.3A.

Step by step solution

01

Definition of power, specific heat of a substance and relation between potential difference & power.

Power: The power of a process is the amount of some type of energy converted into a different type divided by the time intervalฮ”tin which the process occurred:

P=ฮ”Eฮ”t โ€ฆ.. (1)

The SI unit of power is the watt (W).

The energy Q required to change the temperature of a mass m of a substance by an amountฮ”Tis

Q=mcฮ”T โ€ฆ.. (2)

Here, c is the specific heat of the substance.

If a potential differenceฮ”Vis maintained across a circuit element, the power, or rate at which energy is supplied to the element, is

P=Iฮ”V โ€ฆ.. (3)

02

Given data and required data:

The mass of water,m=1500kg.

The temperature difference of the water is:

ฮ”T=40.0โˆ˜Cโˆ’10.0โˆ˜C=30.0โˆ˜C

The efficiency of the energy conversion isฮท=75.0%.

The cost of electricity is9cents/centskWโ‹…hkWโ‹…h.

The voltage across the electric heater,ฮ”V=220V.

The time interval of the heating process is:

ฮ”t=(4.00h)(60min1h)(60s1min)=14400s

The specific heat of water, c=4186J/Jkgโ‹…โˆ˜Ckgโ‹…โˆ˜C

03

(a) Calculation of electrical energy supplied by heater and its cost:

The electrical energyEelectricalsupplied by the electric heater goes into increasing the temperature the water a temperature differenceฮ”T.

But we are given that the efficiency of this energy conversion is75.0%.So

(0.95)Eclectrical=Qwater

Here,Qwateris the energy required to raise the temperature of the water found from Equation (2):

(0.95)Eelectrical=mcฮ”T

Solve the above equation forEelectricalas below.

Eelectrical=mcฮ”T0.95

Entering the values for m, c, andฮ”Tin the above equation, it gives:

Eelectrical=(1500kg)(4186J/Jkgโ‹…โˆ˜Ckgโ‹…โˆ˜C)(30.0โˆ˜C)0.95=251.16ร—106J

Convert this energy into the units of kilowatt-hours by multiplying by the following conversion factor:

Eelectrical=(251.16ร—106J)(1kWร—h3.60ร—106J)=209330kWโ‹…h

The cost of this energy is found by multiplying by the cost of aEelectricalsingle kilowatt-hour are given:

Cost=(209330kWโ‹…h)(9cents1kWโ‹…h)=628cents=(628cents)($1100cents)=$6.28

Therefore, Cost of the energy supplied to heater is $6.28.

04

(b) Calculation of amount of current used by heater:

The electrical energyEelectricalis found in terms of the power output of the heater and the time interval of the heating process by Equation (2):

Eelectrical=Pฮ”t

Substitute for P from equation (3):

Eelectrical=Iฮ”Vฮ”t

Solve the above equation for I as below.

I=Eelectricalฮ”Vฮ”t

Substitute numerical values:

I=(251.16ร—106J)(220V)(14400s)=79.3A

Hence, amount of current used by heater is 79.3A.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free