A body of specific heat capacity 450 Jkg\(^{-1}\)K\(^{-1}\) falls to the ground from rest through a vertical height of 20 m. Assuming conservation of energy, calculate the change in temperature of the body striking the ground level. (g = 10 ms\(^{-2}\))
A
0°C
B
\(\frac{2}{9}\) °C
C
\(\frac{4}{9}\) °C
D
\(\frac{9}{4}\) °C
Verified Correct Answer: C.
Step-by-Step Mathematical Solution
c = 450 Jkg\(^{-1}\)K\(^{-1}\); h = 20 m; θ = ?; g = 10 ms\(^{-2}\)
mc\(\Delta\)\(\theta\) = mgh
c x \(\Delta\)\(\theta\) = g x h
450 x \(\Delta\)\(\theta\) = 20 x 10
\(\Delta\)\(\theta\) = \(\frac{200}{450}\) = \(\frac{20}{45}\)
\(\therefore\) \(\Delta\)\(\theta\) = \(\frac{4}{9}\)ºC
Source: WAEC / JAMB / NECO Classroom Archive (Record #76352)