Finding time to increase temp with power

[crazyog]

Homework Statement

A hot water heater is operated by using solar power. If the solar collector has an area of 6 m^2, and the power per unit area delivered by sunlight is 1000 W/m^2, how long in hours will it take to increase the temperature of 1 m^3 of water from 20C to 60C? (1 cal=4.186 J)

Homework Equations

I'm not really sure...
I know that:
power=(sigma)AeT^4
sigma = Stefan -Boltzmann constant = 5.6696x10^(-8) W/m^2*K^4
A = surface area
e = emissivity
T = surface temp.
but we weren't given emissivity so i don't think I use this...

maybe mc(Tf-Ti)/P = (change in time)

?? I'm really clueless on what equation to use, please help!

 

[Hootenanny]

Homework Statement

A hot water heater is operated by using solar power. If the solar collector has an area of 6 m^2, and the power per unit area delivered by sunlight is 1000 W/m^2, how long in hours will it take to increase the temperature of 1 m^3 of water from 20C to 60C? (1 cal=4.186 J)

Homework Equations

I'm not really sure...
I know that:
power=(sigma)AeT^4
sigma = Stefan -Boltzmann constant = 5.6696x10^(-8) W/m^2*K^4
A = surface area
e = emissivity
T = surface temp.
but we weren't given emissivity so i don't think I use this...

maybe mc(Tf-Ti)/P = (change in time)

?? I'm really clueless on what equation to use, please help!

There is no need to invoke the Stefan-Boltzmann law here.

What is the definition of power?

 

[crazyog]

P = force /area ??

(the answer is stated as 7.8)

 

[Hootenanny]

P = force /area ??

That's pressure, not power.

 

[crazyog]

oh yeah hahahha wow, whoops.

P= Work/change in t
or P = Q/ (delta T)

 

[Hootenanny]

oh yeah hahahha wow, whoops.

P= Work/change in t
or P = Q/ (delta T)

Correct. So you know that power provided and you can work out the work/energy required using one of your aforementioned equations.

 

[crazyog]

So do I use mc(Tf-Ti) to find Q?
(1000 kg)(4186)(60-40)?
I used 1000 kg for the mass because D = m/v
we know the density of water is 10^3 = m/(1m^3)
and solved for m ...= 1000
but this is not giving me 7.8

 

[Hootenanny]

So do I use mc(Tf-Ti) to find Q?
(1000 kg)(4186)(60-40)?

You're good so far. What the next step?

 

[crazyog]

Ok so I did (1000)(4186)(60-40) = 83720000/(1000) = 83720
^ I rearranged the P = Q/ (delta t) solved for delta t = 83720 in secs

then they want it in hours so dived by 3600s (83720)/3600 = 23.25 hrs

the answer is 7.8 according to the book so I am not sure where I am going wrong, please help!

 

[crazyog]

I figure I must be missing something since I am not using the area of 6 m^2

 

[Hootenanny]

I figure I must be missing something since I am not using the area of 6 m^2

Indeed you are:

Ok so I did (1000)(4186)(60-40) = 83720000/(1000) = 83720
^ I rearranged the P = Q/ (delta t) solved for delta t = 83720 in secs

If the solar collector has an area of 6 m^2, and the power per unit area delivered by sunlight is 1000 W/m^2

This means that each square meter on the surface of the Earth receives 1000W of energy from the sun.

 

[crazyog]

oh yeah i see that now
soooo..
P = 1000 W/m^2 and there are 6m^2 so it should be multiplied by 6
1000)(4186)(60-40) = 83720000/(6000) = 13953.3333
and then divide by 3600...3.875
still not 7.8 =/

 

[crazyog]

I realized my mistake!
It should be 60-20

(1000)(4186)(60-20)/(6000)
and then divided by 3600
=7.75

thank you so much for your help!

 

[Hootenanny]

I realized my mistake!
It should be 60-20

(1000)(4186)(60-20)/(6000)
and then divided by 3600
=7.75

thank you so much for your help!

A pleasure