# Total power and specific heat

1. May 6, 2007

### kbyws37

If the total power per unit area from the Sun incident on a horizontal leaf is 9.00x10^2 W/m2, and we assume that 70.0% of this energy goes into heating the leaf, what would be the rate of temperature rise of the leaf? The specific heat of the leaf is 2.80 kJ/(kg°C), the leaf area is 7.00 x10^−3 m2, and its mass is 0.200 g.

I don't know what equation to use.
I tried using

Power = kA (Temp/d)
but did not get the right answer.

The right answer is 7.88 degree celsius per sec.

2. May 6, 2007

### hage567

Are you sure that's the right answer? It seems pretty big (or fast).

3. May 6, 2007

### kbyws37

yup, it says that is the correct answer

4. May 6, 2007

### hage567

Well, I would use Q=mc(deltaT).

The rate of change of thermal energy is the power.

I don't quite get the same answer as the book though, I get 7.88x10^-4 oC/s. (which is much more reasonable, in my opinion)

5. May 6, 2007

### kbyws37

If I were to use the equation
Q=mc(deltaT).

Q = (9.00 x 10^2 W/m2) * 0.7 (would this be right b/c it's 70% of energy?)
m= 0.200 g
c= 2.80 kJ/(kg C)
(delta T) = what we are looking for

Where woul the leaf area fit into the picture?

6. May 7, 2007

### kbyws37

can anyone help please?

7. May 7, 2007

### Hootenanny

Staff Emeritus
Technically, since your working with power your equation should be;

$$\frac{dQ}{dt} = mc\Delta T$$

Now, you've done everything correctly up to now. So, if the sun emits 9.00 x 10^2 watts per square meter; much power falls on a leaf with an area of 7.00 x10^−3 square meters?

8. May 7, 2007

### kbyws37

sorry i don't understand the hints.
i did 7.00 x10^−3 / 9.00 x 10^2 watts for dQ/dt
and plugged in m and c and solved for delta T.
still not getting the right answer

9. May 7, 2007

### Hootenanny

Staff Emeritus
Your very close here;
But why are you dividing the area by the power per unit area when you want power? Lets have a look at the units;

$$area = m^2$$

$$\text{power per unit area}=\frac{W}{m^2}$$

$$\text{power}= \frac{dQ}{dt}=W$$

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