The surface temperature of a swimming pool on a warm day is 25ºC and the temperature at the bottom is 15ºC. If the swimming pool has a surface area of 620 \(m^2\) and a depth of 1.5m. Find the rate at which energy is transferred by conduction from the surface to the bottom of the swimming pool.
[Thermal conductivity of water (k) = 0.6071 Wm-1K-1]
[Thermal conductivity of water (k) = 0.6071 Wm-1K-1]
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Correct Answer: Option A
Explanation:
L=1.5m, A=\(620m^2\), ø=25-15=10°c
K=0.6071 , \(Wm^{-1}k^{-1}\) , \(\frac{q}{t}\) =?
\(\frac{q}{t}\) =KA, \(\frac{â–ºt}{L}\)
where k is the thermal conductivity constant and \(\frac{q}{t}\) is the rate of heat transfer
=\(\frac{q}{t}\)=0.6071×620×\(\frac{10}{1.5}\)
\(\frac{q}{t}\) = 2509.35w ≈ 2.5kw
L=1.5m, A=\(620m^2\), ø=25-15=10°c
K=0.6071 , \(Wm^{-1}k^{-1}\) , \(\frac{q}{t}\) =?
\(\frac{q}{t}\) =KA, \(\frac{â–ºt}{L}\)
where k is the thermal conductivity constant and \(\frac{q}{t}\) is the rate of heat transfer
=\(\frac{q}{t}\)=0.6071×620×\(\frac{10}{1.5}\)
\(\frac{q}{t}\) = 2509.35w ≈ 2.5kw