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Steel ball is dropped from a height of 1 M on two horizontal nonconducting surface. Every time it bounces it reaches 80% of its previous height. Nearly by how much will the temperature of the ball rise after 4 bounces? Specific heat capacity of the ball is 0.1calorie per gram degree Celsius. Neglect loss in heat to the surroundings and the floor

Steel ball is dropped from a height of 1 M on two horizontal nonconducting surface. Every time it bounces it reaches 80% of its previous height. Nearly by how much will the temperature of the ball rise after 4 bounces? Specific heat capacity of the ball is 0.1calorie per gram degree Celsius. Neglect loss in heat to the surroundings and the floor

Grade:10

3 Answers

Arun
25750 Points
6 years ago
\DeltaH = mg × 1 × 0.2 + mg × 1 × 0.8 × 0.2 + mg × 0.8 × 0.8 × 0.2 .....(1)
\DeltaH = m × 500 \DeltaT .....(2)
2 + 1.6 + 1.28 = 500 \DeltaT
\DeltaT = 0.01°C
Sanju
106 Points
6 years ago
Let `m` be the mass of the steel b
Arun
25750 Points
6 years ago
Delta H = mg* 0.1 *0.2 + mg*1 *0.8*0.2 +mg*0.8*0.8*0.2
Delta H = m *500 (delta T)
 
2 + 1.6 + 1.28 = 500 (delta T)
Delta T = 0.01 °c
 
 
Regards
Arun (askIITians forum expert)

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