Simple work and energy problem I have the following problem:

A man that weights 50kg goes up running the stairs of a tower in Chicago that is 443m tall. What is the power measured in watts if he arrives at the top of the tower in 15min?

So far I used this equation:

W = ΔEc + ΔEp + ΔEq

Where ΔEc is the variation of the kinetic energy, ΔEp is the variation of the potential energy (based on the height) and ΔEq is the energy based on friction.
There isn't an initial W and a ΔEq, so I have this equation:

ΔEc = - ΔEq

Am I doing something wrong? I am stuck here.
 A: Assuming man stops on reaching at top, all the energy is now converted into potential energy. I am ignoring friction loss etc as no such information is available. Given this
Potential Energy = mgh 
and Power = Energy/Time
Thus assuming g =9.8m/sec^2
Power = 241.18 Watt
A: Assume he starts off stationary, and ends up stationary (he'll accelerate and decelerate in between, but that's by the by). So now you know ${\Delta}Ec$ .
Now, you say there is no initial $W$, but you don't say what $W$ is. I'm guessing it's work done. (and remember the relationship between work done, power and time). Initial work is meaningless: ${\Delta}Ec + {\Delta}Ep + {\Delta}Eq$ equals total work done, i.e. $W$.
And finally, where does the energy consumed in friction end up? Will that be significant relative to ${\Delta}Ep$?
A: First, we have to get the weight of the person from his mass.  50 kg here on earth has a weight of 490 N.  That force times the distance over which it is applied gives the total energy: 490 N x 443 m = 217 kJ.  That amount of work was done in 15 minutes, which is 900 seconds.  217 kJ / 900 s = 241 W average over the 15 minutes.
Yes, it really is that simple.
