Question

A rock of mass 0.225 kg falls from rest from a height of 19.0 m into a pail containing 0.432 kg of water. The rock and water have the same initial temperature. The specific heat capacity of the rock is 1840 J/kg C°. Ignore the heat absorbed by the pail itself, and determine the rise in temperature of the rock and water in Celsius degrees.

Answer 1

Answer:

rise in temperature of water and rock is 0.01885 °C.

Explanation:

mass of rock, m = 0.225 kg

height, h = 19 m

mass of water = 0.432 kg

specific heat of rock = 1840 J/kg°C

specific heat of water = 4186 J/kg°C

Potential energy of rock = heat gained by the rock + heat gained by the water

let ΔT be the rise in temperature of rock

mass of rock x height x g = mass of rock  x specific heat of rock x ΔT + mass of water x specific heat of water x ΔT

0.225 x 9.8 x 19 = 0.225 x 1840 x ΔT + 0.432 x 4186 x ΔT

41.895 = (414 + 1808.352) x ΔT

Δ T = 0.01885 °C

Thus, the rise in temperature of rock and water is 0.01885 °C.