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\(^{-2}\))
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
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