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A ball dropped from a height of 2.0 m bounces back to 1.5 m after striking the floor. If 40% of the mechanical energy lost is converted into heat retained by the ball, find the rise in the ball's temperature. The specific heat capacity of the ball is 800 J/(kg K).

1 x 10⁻³ deg C
1.5 x 10⁻³ deg C
2.5 x 10⁻³ deg C
3 x 10⁻³ deg C

Correct answer: 2.5 x 10⁻³ deg C

Solution

The mechanical energy lost per kg is g(h1-h2); 40% of this equals c*delta_T, giving delta_T = 2.5 x 10⁻³ deg C.

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