Arock 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.

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.

seeing the graph would

that tool is called an altimeter which measures your altitude which can be used to tell which atmosphere you are currently in, including the stratosphere.

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