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**Tweety** Okay I asked my teacher and he said I have to use this equation for the second part,

thermal resistance for a single pipe = $\displaystyle \frac{ln(\frac{r_{o}}{r_{i}})}{2 \pi H \lambda} $

$\displaystyle H = 4m $

so $\displaystyle r_{o} = 2 cm = 0.02 m $

$\displaystyle r_{i} = 1.5cm = 0.015 m $

$\displaystyle \lambda = 19 W m^{-1} K^{-1} $

pluggiing these numbers in I get $\displaystyle 6.02\times 10^{-4} $

so the overall resistance, I am not sure but Can I just use the same formula as in electrical circuits, e.g $\displaystyle \frac{1}{R} = \frac{1}{R} + \frac{1}{R} ... $

and times that by 50, since there are 50 tubes

so R = $\displaystyle \frac{1}{ 6.02 x 10^{-4}} \times 50 $

Is this correct?

I really need help, with the first part, since we are turning water into steam will I have to use Q = ml , for latent heat? or is it $\displaystyle Q = mc \delta T $.