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]
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option A
Explanation:
mgh = mc\(\theta\)
\(\theta = \frac{gh}{c} = \frac{10 \times 50}{4200}\)
= 0.119oC
mgh = mc\(\theta\)
\(\theta = \frac{gh}{c} = \frac{10 \times 50}{4200}\)
= 0.119oC