A piece of substance of specific heat capacity 450Jkg\(^{-1}\)k\(^{1}\) falls through a vertical distance of 20m from rest. Calculate the rise in temperature of the substance on hitting the ground when all its energies are converted into heat. [g = 10ms\(^{-2}\)]
- \(\frac{2}{9}\)ºC
-
\(\frac{4}{9}\)ºC
- \(\frac{9}{2}\)ºC
- \(\frac{9}{4}\)ºC
The correct answer is: B
Explanation
Using the law of conservation of energy
mc\(\bigtriangleup\theta\) = mgh
\(\bigtriangleup\theta\) = \(\frac{gh}{c} = \frac{10 \times 20}{450} = \frac{4}{9}\)ºC
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