The device shown below contains 2 kg of water. When the cylinder is allowed to fall 250 m, the temperature of the water increases by 1.4°C. Suppose 2 kg of water are added to the container and the cylinder is allowed to fall 750 m. What would the increase in temperature be in this case?

same height ... half the temp. increase ... must be twice the amount of water

Well, if the water is taking a fall, I hope it remembers to tuck and roll! Anyway, let's calculate the increase in temperature.

We know that the change in temperature is proportional to the distance fallen. So, in the first case, if a 250 m fall resulted in a 1.4°C increase, we can set up a proportion:

250m / 1.4°C = 750m / x

Cross-multiplying, we get: 250m * x = 1.4°C * 750m
Rearranging, we find: x = (1.4°C * 750m) / 250m

Simplifying, we get x = 4.2°C

So, in this case, the increase in temperature would be 4.2°C. Hopefully, the water enjoys its plummeting adventures!

To answer this question, we can make use of the concept of potential energy and the specific heat capacity of water.

First, let's consider the initial situation where the cylinder is falling 250 m and the temperature of the water increases by 1.4°C. We know that as the cylinder falls, potential energy is converted into thermal energy, leading to a change in temperature.

To find the change in temperature, we'll use the formula:

ΔT = ΔPE / (m * c)

Where:
ΔT is the change in temperature
ΔPE is the change in potential energy
m is the mass of the water
c is the specific heat capacity of water

Given that the cylinder falls 250 m, we can calculate the change in potential energy:

ΔPE = m * g * h

Where:
m is the mass of the water
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the height or distance the cylinder falls

Let's substitute the values:

ΔPE = 2 kg * 9.8 m/s^2 * 250 m
ΔPE = 4900 J

Now we can substitute the values into the initial formula to find the change in temperature:

ΔT = 4900 J / (2 kg * c)

The specific heat capacity of water is approximately 4186 J/kg°C. Plugging in this value and solving for ΔT, we get:

ΔT = 4900 J / (2 kg * 4186 J/kg°C)
ΔT ≈ 0.586 °C

So, in the initial case where the cylinder falls 250 m, the temperature of the water increases by approximately 0.586°C.

Now let's move on to the second scenario where 2 kg of water are added to the container and the cylinder falls 750 m. We want to find the increase in temperature in this case.

Again, we can calculate the change in potential energy:

ΔPE = m * g * h
ΔPE = 4 kg * 9.8 m/s^2 * 750 m
ΔPE = 29400 J

Using the same formula as before, we can find the change in temperature:

ΔT = 29400 J / (4 kg * c)
ΔT ≈ 1.398 °C

Therefore, in the second case where 2 kg of water are added and the cylinder falls 750 m, the increase in temperature would be approximately 1.398°C.

work (energy) is proportional to the height

doubling the amount of water will halve the temperature change

Δt = 1.4ºC * (750 m / 250 m) * [2 kg / (2 kg + 2 kg)]

The device shown below contains 2 kg of water. The cylinder is allowed to fall 800 m during which the temperature of the water increases by 2.4°C. Some amount of water is added to the container and the experiment is repeated. All other values remain constant. This time the temperature of the water increases by 1.2°C. How much water was added to the container?