||| - Official Maths / Science Thread - |||

[Bizzare]

(2(5)^2)k please

I canna calculate eet

[Ronaldo95163]

BTW guys what's your favourite topic in Maths?

[Corn Bird]

Time for some Physics :p

Air passes over an electrical heater at steady rate of 2500cm cubed per second??. The steady inlet temperature of the air is 20 degrees celsius and the steady outlet temperature is 40 degrees celsius.

What heat is absorbed by air passing over the heater in two hours?

Obtain an estimate for the power rating of the heater.

Density of air: 1.2kg/m cubed
specific heat capacity of air: 1000J/kgK

my physics is rusty. in one second,

heat absorbed=mass x specific heat capacity x temp change
=(density x volume) x specific heat capacity x temp change
=(1.2 x 2500/10^6) x 1000 x 20

this should be equal to the power rating of the heater



heat absorbed in two hours= (heat absorbed in one second) x 7200

[Corn Bird]

gracen, have you seen the math video lectures at

http://www.khanacademy.org/

[Ronaldo95163]

gracen, have you seen the math video lectures at

http://www.khanacademy.org/

Yup...very good stuff there. PatrickJMT is great as well

As for the physics..the Answer for the first part is 4.32x10^5 and the second part is 60J...I don't have my scientific calculator at hand so I don't know what answer you got lol

[Ronaldo95163]

Rate:2500cm^3
Temperature Change: Theta2-Theta1 = 40-20 = 2-
Density of Air= 1.2kgm^3

Heat absorbed:

Eh = mc(deltatheta)

Mass
D=M/V
1.2=x/2.5x10^-3
x=(1.2)(2.5x10^-3)
x=3x10^-3

Eh= 3x10^-3x1000x20
=60J

In 2 hours

2hrs->s =
2*3600=7200s

60*7200 = 432000
=4.32x10^5

Power = Work done/Time taken
Work done = energy transformed
Work Done = 4.32x10^5
=>Power= 4.32x10^5/7200
=60J

Your answers were correct as well.

[Corn Bird]

Rate:2500cm^3


=>Power= 4.32x10^5/7200
=60J

Your answers were correct as well.

might want to keep an eye on the units. i guess physics teachers don't beat anymore

[Ronaldo95163]

Rate:2500cm^3


=>Power= 4.32x10^5/7200
=60J

Your answers were correct as well.

might want to keep an eye on the units. i guess physics teachers don't beat anymore

J/s sorry


What's wrong with the centimeter cubed?
What's wrong with the cm^3?

2500cm^3 is the Volume as well


BTW I think I fell in love

[DEVIANT]

aye the quadratic equation.....btw, we physics teachers can't beat the children because they will bring police for us....

[Corn Bird]

aye the quadratic equation.....btw, we physics teachers can't beat the children because they will bring police for us....

ok boss. without violence, tell gracen that 'rate' usually means change of some physical quantity with time. for example

velocity=rate of change of distance

will have unit m/s

Air passes over an electrical heater at steady rate of 2500cm cubed. this should be 2500 cm^3/s

also J/s is referred to as W (Watts)

[Ronaldo95163]

oooh yeah

Thanks man

[nervewrecker]

Time for some Physics :p

Air passes over an electrical heater at steady rate of 2500cm cubed per second??. The steady inlet temperature of the air is 20 degrees celsius and the steady outlet temperature is 40 degrees celsius.

What heat is absorbed by air passing over the heater in two hours?

Obtain an estimate for the power rating of the heater.

Density of air: 1.2kg/m cubed
specific heat capacity of air: 1000J/kgK

Volume of air that passes through per second = 0.000250m^3

Mass of air that passes through per second = 0.000250/1.2 kg = 0.000208 kg

temp rise = 40K -20K = 20K

shc of air = 1kJ / kgK

density of air = 1.2kg / m^3

-----------------------------------------------------------------------------

I donno if I approaching this the wrong way but:

shouldnt it take 0.000208 * 1kJ to raise 0.000208 kg of air 1 K (0.208J)

& since it raises it 20K it should be 0.208 * 20 = 4.16 J.

Amount of heat absorbed in 2 hours = 4.16 * 7200 = 29.95kJ

----------------------------------------------------------------------------
Power rating for heater

J = watts / second

Heater puts out 4.16J or 4.16 watts per second.

[nervewrecker]

steups, my first figure is wrong, its 0.002500m^3 not 0.000250m^3

therefore the mass should be 0.00208kg

& 1.2kg / 1000 = the amount of air 1J raises by 1K = 0.0012kg

0.00208 / 0.0012 = 1.73

1.73 that amount passes through. It will take 1.73J to raise 0.00208kg of air 1 degree kelvin 1.73 * 20 the amount it will take to raise it by 20 degrees = 11.56 J.

[Ronaldo95163]

Na the answers are 4.32x10^5 and 60J/s

[Ronaldo95163]

Found out that yesterday was World Maths day

[nervewrecker]

well I miscalculate somwhere then, I will check over when I get more time.

[Ronaldo95163]

yea...you were supposed to get the mass as 3x10^-3kg

[nervewrecker]

2500cc = 0.0025 of 1m^3 which should be 0.0025 * 1.2kg for the weight as well which will be 0.003kg.

0.003 kg of air passes over the heater per second. I need coffee to do this yes.

[Ronaldo95163]

Getting the answer isn't everything though....once you understand the method it doesn't really matter....but getting the answer would give full marks


BTW I MUST HAVE THAT CALCULATOR UGHHHHHHHH!

[Corn Bird]

BTW I MUST HAVE THAT CALCULATOR UGHHHHHHHH!

how about some math software that can do much more than this calculator can?

http://www.sagemath.org/tour-graphics.html

if you know someone at uwi, perhaps they can get you a copy of Maple

http://www.maplesoft.com/products/Maple/features/

[Ronaldo95163]

BTW I MUST HAVE THAT CALCULATOR UGHHHHHHHH!

how about some math software that can do much more than this calculator can?

http://www.sagemath.org/tour-graphics.html

if you know someone at uwi, perhaps they can get you a copy of Maple

http://www.maplesoft.com/products/Maple/features/

The math software is cool but I wanted something a bit more advanced for school..

Right now i'm using this:



Clear Blue version though

[black start]

^them calcs does fail to quickly, I much prefer the casio....

liking the thermo questions doh....

[Ronaldo95163]

^Had mine for about 4 years now....and the screen fades ever so often....changed the batteries and I still get the same issue...so i'm looking to upgrade soon.

[Ronaldo95163]

How ched ded so?

Some more questions will arise soon

[stev]

just a lil humor:




now, ah ketchin mih arse wit dis:

A man wanted to get into his work building, but he had forgotten his code. However, he did remember five clues. These are what those clues were:

The fifth number plus the third number equals fourteen.

The fourth number is one more than the second number.

The first number is one less than twice the second number.

The second number plus the third number equals ten.

The sum of all five numbers is 30.

What were the five numbers and in what order?

[Ronaldo95163]

:lol: just a lil humor:




now, ah ketchin mih arse wit dis:

A man wanted to get into his work building, but he had forgotten his code. However, he did remember five clues. These are what those clues were:

The fifth number plus the third number equals fourteen.

The fourth number is one more than the second number.

The first number is one less than twice the second number.

The second number plus the third number equals ten.

The sum of all five numbers is 30.

What were the five numbers and in what order?

Series question....

I go try eet

[Ronaldo95163]

Not series....not working out like that.... saw the sum of the 5 numbers is 30 and what popped into my head was S5=30...

try giving the numbers variable names and form equations from the clues

[Corn Bird]

try giving the numbers variable names and form equations from the clues

yes, this is how to do it.

x5+x3=14
x4-x2=1
2x2-x1=1
x2+x3=10
x1+x2+x3+x4+x5=30

the problem is the actual calculation, that is, solving the resulting 5 x 5 system by hand. idea is to do it systematically using a technique from linear algebra called row reduction

write equations in echelon form as much as possible:

x1 + x2+ x3+ x4+ x5 =30
-x1+2x2+ 0 + 0 + 0 =1
0 + x2 + x3+ 0 + 0 =10
0 - x2 + 0 + x4 + 0 =1
0 + 0 + x3 + 0 +x5 =14

then put into augmented matrix form

| 1 1 1 1 1 30 |
| 0 -1 0 1 0 1 |
| 0 1 1 0 0 10 |
| 0 -1 0 1 0 1 |
| 0 0 1 0 1 14 |

then use row reduction to get the answer:

x1=7 x2=4 x3=6 x4=5 x5=8

[Ronaldo95163]

Nice!!!!

[stev]

corn bird buss up d question like ah boss.