WAEC - Physics (2005 - No. 20)
Water falls through a height of 50m. Determine the temperature rise at the bottom of the fall. [Neglect energy losses. Specific heat capacity of water = 4200Jkg-1 K-1, g = 10ms2]
0.119oC
0.840oC
1.190oC
8.400oC
Explanation
mgh = mc\(\theta\)
\(\theta = \frac{gh}{c} = \frac{10 \times 50}{4200}\)
= 0.119oC
\(\theta = \frac{gh}{c} = \frac{10 \times 50}{4200}\)
= 0.119oC
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