In Joule's experiments, the slow lowering of a weight (through a pulley and cable arrangement) turned a stir - rer in an insulated container of water. As a result of viscosity, the kinetic energy transferred from the stirrer to the water eventually dissipated. In this process the po - tential energy of the weight was first converted to kinetic energy of the stirrer and the water, and then as a result of viscous forces, the kinetic energy of the water was converted to thermal energy apparent as a rise in tem - perature. Assuming no friction in the pulleys and no heat losses, how large a temperature rise would be found in 1 kg of water as a result of a 1-kg weight being lowered 1m?

Question

Cecilia Adkins

Chemistry

16 December, 16:03

In Joule's experiments, the slow lowering of a weight (through a pulley and cable arrangement) turned a stir - rer in an insulated container of water. As a result of viscosity, the kinetic energy transferred from the stirrer to the water eventually dissipated. In this process the po - tential energy of the weight was first converted to kinetic energy of the stirrer and the water, and then as a result of viscous forces, the kinetic energy of the water was converted to thermal energy apparent as a rise in tem - perature. Assuming no friction in the pulleys and no heat losses, how large a temperature rise would be found in 1 kg of water as a result of a 1-kg weight being lowered 1m?

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Answers (1)

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Izabelle Kirby · Answer 1

The temperature rise will be 0.0023 °C

Explanation:

We know that:

Potential Energy = mgh

Kinetic Energy = (1/2) mv²

Thermal Energy = mCΔT

where,

m = mass = 1 kg

h = height = 1 m

v = velocity

C = specific heat of water = 4186 J/kg °C

ΔT = rise in temperature

According to given situation:

Potential Energy of Weight = Kinetic Energy of Stirrer and water

Kinetic energy of stirrer and water = Thermal Energy of water

Therefore, it implies that:

Potential Energy of Weight = Thermal Energy of water

mgh = mCΔT

ΔT = gh/C

ΔT = (9.8 m/s²) (1 m) / (4186 J/kg °C)

ΔT = 0.0023 °C