Question

The apparatus shown in the figure below was used by Joule to measure the mechanical equivalent of heat. Work is done on the water by a rotating paddle wheel, which is driven by two blocks falling at a constant speed. The temperature of the stirred water increases due to friction between the water and the paddles. If the energy lost in the bearings and through the walls is neglected, then the loss in potential energy associated with the blocks equals the work done by the paddle wheel on the water. If each block has a mass of 1.30 kg and the insulated tank is filled with 200 g of water, what is the increase in temperature of the water after the blocks fall through a distance of 2.75 m? Answer is in °C.

Answer 1

The increase in temperature of the water after the blocks fall through a distance of 2.75 m is approximately 8.4 °C.

To find the increase in temperature of the water, we can use the formula:

Q = mcΔT

where Q is the heat transferred, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

The heat transferred to the water is equal to the work done by the paddle wheel on the water, which is equal to the loss in potential energy associated with the blocks:

Q = mgh

where m is the mass of each block, g is the acceleration due to gravity, and h is the height through which the blocks fall.

Substituting in the given values:

Q = 2mgh = 2(1.30 kg)(9.81 m/s^2)(2.75 m) = 70.6 J

The specific heat capacity of water is 4.18 J/g·°C, so:

70.6 J = (0.200 kg)(4.18 J/g·°C)(ΔT)

Solving for ΔT:

ΔT = 70.6 J / (0.200 kg)(4.18 J/g·°C) ≈ 8.4 °C

Therefore, the increase in temperature of the water after the blocks fall through a distance of 2.75 m is approximately 8.4 °C.

Answer 2

Answer:Here are the steps to solve this problem:

Calculate the potential energy lost by the blocks:

PE = mgh

PE = (1.3 kg)(9.8 m/s^2)(2.75 m) = 37.7 J

The potential energy lost equals the work done by the paddles on the water:

W = PE = 37.7 J

Calculate the mass of the water:

Given: Mass of water = 200 g = 0.2 kg

Calculate the specific heat capacity of water:

cp = q/mΔT

q = W = 37.7 J (heat added)

m = 0.2 kg (mass of water)

cp = 4187 J/kg•°C (specific heat capacity of water)

Solve for the change in temperature:

ΔT = q/mcp

ΔT= 37.7 J / (0.2 kg * 4187 J/kg•°C)

ΔT = 0.0045°C

Therefore, the increase in temperature of the water after the blocks fall is 0.0045°C.

In summary, by applying the work-energy principle (potential energy lost equals work done), conservation of energy, and using properties of water, we calculated the temperature increase of the stirred water to be 0.0045°C.

Step-by-step explanation: