A lead ball, with an initial temperature of 25.0 deg. C, is released from a height of 115.0 m. It does not bounce when it hits a hard surface. Assume all the energy of the fall goes into heating the lead. Find the temperature (in degrees C) of the ball after it hits. Data: clead = 128 J/kgoC.
2007-04-28 04:37:42 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
This is just a matter of converting the initial gravitational potential energy into thermal energy. To get the temperature rise, divide the energy by the specific heat: ΔT = g·h/cp. You don't need to know the mass; g·h is the specific energy, that is, energy per unit mass.
2007-04-28 05:04:14 · answer #1 · answered by injanier 7 · 1⤊ 0⤋
You don not need to know the mass of the ball: a ball 10 times heavier than another ball will convert 10 times the kinetic energy into heat energy - but over ten times the mass.
Assume mass = 1kg.
Potential energy = kinetic energy at impact = 1 x 115 x 9.8 = 1127J.
Divide this by 128 . . . gives 8.8 deg C so the ball heats up to 33.8 deg.
If you're told to use 10 instead of 9.81, then it works out at 25 + 9 = 34 deg C.
2007-04-28 11:56:20 · answer #2 · answered by JJ 7 · 0⤊ 0⤋
You need to know the mass of the ball to solve this problem
2007-04-28 11:46:31 · answer #3 · answered by Ben 1 · 0⤊ 0⤋