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JAMB Physics 2023 Paper

Question : 13

Total: 65

m^2

[Thermal conductivity of water (k) = 0.6071 Wm-1K-1]

Solution:

L=1.5m, A= $620m^2$ , ø=25-15=10°c

K=0.6071 , $\mathrm{W}\,\mathrm{m}^{-1}\,\mathrm{k}^{-1}$ , $\frac{q}{t}$ =?

\frac{q}{t}

=KA,

\frac{\blacktriangleright t}{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

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