A rock of mass 0.234 kg falls from rest from a height of 28.9 m into a pail containing 0.492 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.

To solve this problem, we need to calculate the amount of heat energy gained by the rock as it falls and then use this energy to determine the rise in temperature of the rock and the water.

Let's start by calculating the potential energy of the rock before it falls:

Potential energy, PE = mass * gravity * height

Given:
Mass of the rock, m_rock = 0.234 kg
Height, h = 28.9 m
Acceleration due to gravity, g = 9.8 m/s^2

PE_rock = m_rock * g * h

Substituting the values:

PE_rock = 0.234 kg * 9.8 m/s^2 * 28.9 m
PE_rock = 65.9172 J

Since the rock falls from rest, all of its potential energy is converted to heat energy when it falls into the water.

Now, let's calculate the heat energy gained by the rock:

Heat energy gained by the rock, Q_rock = PE_rock

Q_rock = 65.9172 J

The heat energy gained by the rock is given by:
Q_rock = m_rock * specific heat capacity_rock * change in temperature_rock

Rearranging the equation, we can find the change in temperature of the rock:

change in temperature_rock = Q_rock / (m_rock * specific heat capacity_rock)

Given:
Specific heat capacity of the rock, specific heat capacity_rock = 1840 J/kg°C

change in temperature_rock = 65.9172 J / (0.234 kg * 1840 J/kg°C)
change in temperature_rock = 0.151°C

Therefore, the temperature of the rock rises by approximately 0.151°C as it falls into the water.

Now, let's calculate the change in temperature of the water:

The heat energy gained by the water can be calculated as the heat energy lost by the rock:

Q_water = -Q_rock
Q_water = -65.9172 J

Using the same equation as before:

Q_water = m_water * specific heat capacity_water * change in temperature_water

Rearranging the equation, we can find the change in temperature of the water:

change in temperature_water = Q_water / (m_water * specific heat capacity_water)

Given:
Mass of the water, m_water = 0.492 kg
Specific heat capacity of water, specific heat capacity_water = 4186 J/kg°C (approximately)

change in temperature_water = -65.9172 J / (0.492 kg * 4186 J/kg°C)
change in temperature_water = -0.031°C

Therefore, the temperature of the water decreases by approximately 0.031°C when the rock falls into it.

Note that the negative sign indicates a decrease in temperature.

To determine the rise in temperature of the rock and water, we can use the equation:

q = m * c * ΔT

Where:
q is the heat absorbed or released by an object (in this case, the rock and water)
m is the mass of the object
c is the specific heat capacity of the material
ΔT is the change in temperature

For the rock:
The heat absorbed by the rock can be calculated using the equation:

q_rock = m_rock * c_rock * ΔT_rock

For the water:
The heat absorbed by the water can be calculated using the equation:

q_water = m_water * c_water * ΔT_water

Since the rock and water have the same initial temperature, the heat absorbed by the rock is equal to the heat released by the water, so:

q_rock = -q_water

Now let's calculate each component step by step.

1. Calculate the heat absorbed by the rock (q_rock):
Using the equation:

q_rock = m_rock * c_rock * ΔT_rock

Given:
m_rock = 0.234 kg
c_rock = 1840 J/kg C°

We need to find ΔT_rock.

2. Calculate the heat released by the water (q_water):
Using the equation:

q_water = m_water * c_water * ΔT_water

Given:
m_water = 0.492 kg
c_water = 4186 J/kg C°

We need to find ΔT_water.

3. Equate q_rock to -q_water and solve for ΔT_rock and ΔT_water.