Question

A 100 kg marble slab falls off a skyscraper and falls 200 m to the ground without hitting anyone. Its fall stops within milliseconds, so that there is no loss of thermal energy to its surroundings if its temperature is measured immediately after it stops. By how much has its temperature changed as a result of the fall, if we ignore energy gained or lost as a result of its interaction with the atmosphere? Cmarble = 860 J/(kg oC) 0.57 °C 1.14 °C 2.28 °C 4.56 °C

Answer 1

Answer:

Δ T = 2.28°C

Explanation:

given,

mass of marble = 100 Kg

height of fall = 200 m

acceleration due to gravity = 9.8 m/s²

C_marble = 860 J/(kg °C)

using conservation of energy

Potential energy = heat energy

  $m g h = m C_{marble}\Delta T$

  $g h =C_{marble}\Delta T$

  $\Delta T= \dfrac{g h}{C_{marble}}$

  $\Delta T= \dfrac{9.8 \times 200}{860}$

        Δ T = 2.28°C