Please, I need help with the 3 questions. I have tried solving them but the result is not consistent with other things im finding online... How do I even calculate the net on the first question... 1)Consider a domestic hot water tank that has a total surface area of A = 2.9 m2. The tank and its contents are maintained at a constant temperature of 58 °C by an electric immersion heater and the temperature of the surroundings outside the tank is 19°C. The emissivity of the tank is ε = 0.76. Calculate the net power radiated by the tank (i.e. the difference between the power radiated and the power absorbed). Give your final answer to an appropriate number of significant figures. 2)The tank is now completely surrounded by an insulating layer of rock mineral wool of thickness l = 0.115 m. The thermal conductivity of rock mineral wool is k = 0.032 W m−1 K−1.You may assume that the surface area of the layer of rock mineral wool is the same as the surface area of the tank. Calculate the power lost through the insulating layer. Give your final answer to an appropriate number of significant figures.

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
icon
Related questions
Question

Please, I need help with the 3 questions. I have tried solving them but the result is not consistent with other things im finding online... How do I even calculate the net on the first question...

1)Consider a domestic hot water tank that has a total surface area of A = 2.9 m2. The tank and its contents are maintained at a constant temperature of 58 °C by an electric immersion heater and the temperature of the surroundings outside the tank is 19°C. The emissivity of the tank is ε = 0.76. Calculate the net power radiated by the tank (i.e. the difference between the power radiated and the power absorbed). Give your final answer to an appropriate number of significant figures.


2)The tank is now completely surrounded by an insulating layer of rock mineral wool of thickness l = 0.115 m. The thermal conductivity of rock mineral wool is k = 0.032 W m−1 K−1.You may assume that the surface area of the layer of rock mineral wool is the same as the surface area of the tank. Calculate the power lost through the insulating layer. Give your final answer to an appropriate number of significant figures.


​​​​​​​3)If the tank is operated continuously, how much money is saved over the course of a year (to the nearest pound), as a result of insulating the tank? (Assume the cost of 1 kWh of electricity is 17.2 pence.)

TOCK
TOC
W
BIG
BIGSTOCK
BIGS
BIG
BIG
PLEASE HELP
BOD BLESS
BIG
BIGST
BIG
BIGSTOCK
BIGSTOCK
Image ID: 19473782
bigstock.com
Transcribed Image Text:TOCK TOC W BIG BIGSTOCK BIGS BIG BIG PLEASE HELP BOD BLESS BIG BIGST BIG BIGSTOCK BIGSTOCK Image ID: 19473782 bigstock.com
Expert Solution
steps

Step by step

Solved in 4 steps with 1 images

Blurred answer
Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question

You just copied the same answer from Chegg but that one is incorrect... so far these are the results i got but i also think they're not correct. 

 

4(a) 

 

The power radiated by the tank can be expressed in the following formula:

P = εσA(Tin4 – Tout4

Where,

ε = 0.76

σ = 5.67 x 10^-8 W/m2K^4

A = 2.8 m^2

T(Tin^4 – Tout^4) = (58 + 273.15 = 331.15 K)^4 – (19 + 273.15 = 292.15 K)^4 = 4740486867.42 K

 

So Pnet= 0.76 * 5.67 × 10^-8 * 2.9 * 4740486867.42

Pnet = 592.4 W or 0.59 kW

 

 

4(b)

 

The power lost through the insulating layer can be calculated using the following formula: 

Ploss = kA Ttan⁡k−Ts / l

where Ploss is the power lost through the insulating layer, k is the thermal conductivity of the rock mineral wool, A is the surface area of the tank, Ttan⁡k is the temperature of the tank, Ts is the temperature of the surroundings, and l is the thickness of the insulating layer.

 

k = 0.032 

A = 2.9 m^2

Ttan⁡k = 58 °C

Ts = 19°C

l= 0.115 m

 

Ploss = 0.032 * 2.9 * 58−19/115

Ploss = 31.47 W

 

4(c)

 

This is the power that the immersion heater must provide to maintain the temperature of the tank and its contents when insulated.

The amount of energy saved is: 

 

ΔE = (Ploss - Pnet) × t 

 

Where t is the time the tank is operated continuously in a year.

Assuming a year has 365 days, we have:  t = 365 days × 24 hours/day × 3600 s/hour = 31536000 s 

 

So,  ΔE = (0.59 kW - 0.03 kW) × 31536000 s = 17660160 kJ 

 

To convert this to kWh, we divide by 3.6 × 10^6 (1 kWh = 3.6 × 10^6 J)

ΔE = 17660160 kJ / (3.6 × 10^6 J/kWh) = 4.9 kWh 

 

Finally, to calculate the money saved, we multiply by the cost of 1 kWh of electricity: 

 

Money saved = 4.9 kWh × 17.2 p/kWh = 84,28 p / 100 =  £0,84

 

Therefore, the money saved over the course of a year as a result of insulating the tank is approximately £0,84.

 

Solution
Bartleby Expert
SEE SOLUTION
Knowledge Booster
Thermal expansion
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON