can someone explain this question?

[voxxine]

Physics IGCSE:
"During one day, 250 kg of water is pumped through the solar panel. The temperature of this
water rises from 16 °C to 38 °C.
The water absorbs 25% of the energy falling on the solar panel, and the specific heat capacity
of water is 4200 J / (kg °C).
Calculate the energy falling on the solar panel during that day."

I've tried and the answer I got is 2.31 x 10^7 J. but it's wrong!!! can someone help??

 

[Z.A.M]

What was the answer?

 

[Z.A.M]

Was the answer 9.24x10^7?

 

[leosco1995]

I think you just calculated the energy that the water absorbs on the water panel. Since that's only 25%, you have to multiply your answer by 4.

Do correct me if I'm wrong though.

 

[Z.A.M]

I think you just calculated the energy that the water absorbs on the water panel. Since that's only 25%, you have to multiply your answer by 4.

Do correct me if I'm wrong though.

R u talking about My anwer? Or voxxines

 

[leosco1995]

R u talking about My anwer? Or voxxines

voxxines. Your answer is right.

 

[Z.A.M]

Okieees

 

[voxxine]

I think you just calculated the energy that the water absorbs on the water panel. Since that's only 25%, you have to multiply your answer by 4.

Do correct me if I'm wrong though.

ok. 25%. but why 4?? 100/25 = 4 ?

 

[Z.A.M]

Look dude the energy used by the water 1/4th or 25% of the total energy so the total energy can be calculated like this:
Q= m.c.change in temp
0.25Q=250x4200x22
0.25Q=2.31x10^7
Q=(2.3x10^7/0.25)
Q=9.24x10^7
I hope u get it dis way..

 

[IGCSE O/L student]

Physics IGCSE:
"During one day, 250 kg of water is pumped through the solar panel. The temperature of this
water rises from 16 °C to 38 °C.
The water absorbs 25% of the energy falling on the solar panel, and the specific heat capacity
of water is 4200 J / (kg °C).
Calculate the energy falling on the solar panel during that day."

I've tried and the answer I got is 2.31 x 10^7 J. but it's wrong!!! can someone help??

is da ans 9.24 x 10^7. awwwww, dont cry!!! cheer up!

 

[voxxine]

Look dude the energy used by the water 1/4th or 25% of the total energy so the total energy can be calculated like this:
Q= m.c.change in temp
0.25Q=250x4200x22
0.25Q=2.31x10^7
Q=(2.3x10^7/0.25)
Q=9.24x10^7
I hope u get it dis way..

thank youuu so much Z.A.M. i get it now

 

[voxxine]

is da ans 9.24 x 10^7. awwwww, dont cry!!! cheer up!

I get it now. thanks

 

[IGCSE O/L student]

I get it now. thanks

ur welcum. if u need help, jst ask me,k !