Find Tank Wall Temp: Heat Conductivity & Loss Analysis

[TSN79]

A tank holds water at 60 degrees C. The outside temp. is 10. The thickness of the walls is 12mm. My goal is to find the temp on the wall on the outside of the tank. I've got some values I don't know the english name for, but hopefully you'll understand by the units.
\lambda for steel is 50\frac{W}{mK}
h between water and tank is 2700\frac{W}{m^2K}
h between air and tank is 15\frac{W}{m^2K}

How do I go about it?? I also need to find the effect loss...

 

[Gokul43201]

How to go about it : The heat flowing per unit area from the water, Q, is equal to the heat received per unit area by the outside air (and is also equal to the heat conducted per unit area through the tank wall). Calculate each of these heat flows and equate them to each other.

I've already said too much... in the future, if you want help, you must show what you've tried and where you are stuck.

 

[TSN79]

First I found the value which we in norwegian denote U [W/(m^2*K)]
U=\frac{1}{(1/h_1)+(\delta_1/\lambda_1)+(1/h_2)}
Putting in the values this gives me U=14,86 W/(m^2*K)
Then I want to use the equation
\dot{Q}=U \cdot A(T_1-T_2)
where A is the area of the tank wall, T1 and T2 are the temp. of the water and the air and Q is the energy that passes through the wall. The problem is that I don't have the area, I only have the thickness of the wall, 0,012m. If I did know A, then I could calculate Q, and use this equation
\dot{Q}=h_1 \cdot A(T_1-T_w_1)
to find T_w_1 which is the temp. at the outside of the tank wall.
The funny thing is that if I follow this procedure and use my value for thickness istead of area, the answer comes out right (!), though the unit of the answer isn't correct. By the way, the answer is 59,5 degrees C.
What am doing wrong? Is there some other equation I should be using?