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At a power station, heat is removed from the heat exchanger by cooling water at $6.7 × 10^9\ J$ per minute. The cooling water enters at $6.0\ ^oC$ and leaves at $14.0\ ^oC$ . [Take the specific heat capacity of water $4200\ J/kg^oC$ ] Which of the following is the rate of water flow?
$\frac{{6.7 \times {{10}^9} \times 60}}{{4200 \times 8}}\ kg\ s^{-1}$
$\frac{{6.7 \times {{10}^9}}}{{4200 \times 8 \times 60}}\ kg\ s^{-1}$
$\frac{{4200 \times 8}}{{6.7 \times {{10}^9} \times 60}}\ kg\ s^{-1}$
$\frac{{4200 \times 8 \times 60}}{{6.7 \times {{10}^9}}}\ kg\ s^{-1}$
Solution
$\frac{\mathrm{d} \theta}{\mathrm{dt}}=\frac{\mathrm{d} \mathrm{m}}{\mathrm{dt}} \times \mathrm{S} \times \mathrm{dT}$
$\Rightarrow \frac{6.7 \times 10^{9}}{60}=\frac{\mathrm{dm}}{\mathrm{dt}} \times 4200 \times(14-6)$
$\Rightarrow \frac{d m}{d t}=\frac{6.7 \times 10^{9}}{4200 \times 8 \times 60}$
