Problem 91 - Physics

Question

A rock of mass 0.20 kg falls from rest from a height of \(15 \mathrm{~m}\) into a pail containing \(0.35 \mathrm{~kg}\) of water. The rock and water have the same initial temperature. The specific heat capacity of the rock is \(1840 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\). Ignore the heat absorbed by the pail itself, and determine the rise in the temperature of the rock and water.

Step-by-Step Solution

Answer

The temperature rises by approximately 0.015°C.

Step 1: Calculate Potential Energy

First, calculate the gravitational potential energy (GPE) of the rock before it falls using the formula \( GPE = mgh \), where \( m = 0.20 \) kg, \( g = 9.81 \) m/s² (acceleration due to gravity), and \( h = 15 \) m.\[ GPE = 0.20 \times 9.81 \times 15 = 29.43 \text{ J} \]

Step 2: Apply Energy Conservation

Assuming no energy loss (i.e., via sound, air resistance), the potential energy converts into thermal energy, leading to a temperature increase of the rock and water.

Step 3: Determine Total Mass and Specific Heat

Calculate the total mass of the rock and water, \( m_{total} = 0.20 + 0.35 = 0.55 \) kg. The specific heat capacity \( c_{eq} \) can be calculated using the formula for average specific heat:\[ c_{eq} = \frac{m_{rock}c_{rock} + m_{water}c_{water}}{m_{total}} \]With specific heat of water \( c_{water} = 4186\, \mathrm{J}/(\mathrm{kg} \cdot \mathrm{C}^{\circ}) \), we have:\[ c_{eq} = \frac{0.20 \times 1840 + 0.35 \times 4186}{0.55} \approx 3542.73 \, \mathrm{J}/(\mathrm{kg} \cdot \mathrm{C}^{\circ}) \]

Step 4: Calculate Temperature Rise

Use the formula for heat transfer, \( Q = m_{total} c_{eq} \Delta T \). Here, \( Q = 29.43 \) J, \( m_{total} = 0.55 \) kg, and \( c_{eq} = 3542.73 \, \mathrm{J}/(\mathrm{kg} \cdot \mathrm{C}^{\circ}) \). Solve for \( \Delta T \):\[ \Delta T = \frac{Q}{m_{total} c_{eq}} = \frac{29.43}{0.55 \times 3542.73} \approx 0.015 \underline{\phantom{xxx}} \mathrm{C}^{\circ} \]

Step 5: Conclusion

The rise in temperature of the rock and water when the rock falls into the pail is approximately \( 0.015 \, \mathrm{C}^{\circ} \).

Key Concepts

Potential energy

Potential energy is a crucial concept in understanding how energy is stored within an object due to its position relative to other objects. In terms of gravitational potential energy, which is what we see in exercises involving heights and masses, this is the energy an object has due to its position in a gravitational field.

The formula used to calculate gravitational potential energy is \[ GPE = mgh \]where:

• \( m \) is the mass of the object (in kilograms),
• \( g \) is the acceleration due to gravity (usually approximated as \( 9.81 \, \text{m/s}^2 \)),
• \( h \) is the height above the ground (in meters).

This formula helps us calculate how much energy is stored in an object due to its height, which can be converted into other forms of energy, such as kinetic energy or thermal energy, when the object moves or falls.

Heat calculation

Heat calculation involves determining how much thermal energy is involved when there is a change in temperature or phase of a substance. This type of calculation is vital in thermodynamics, as it allows us to understand and predict how substances will behave under different thermal conditions.

In the given exercise, the gravitational potential energy of a falling rock is converted to heat energy upon impact with water. This highlights the principle of energy conservation, where energy changes form but is not lost. This conversion is used to calculate the rise in temperature, as the initial potential energy becomes heat.

The specific formula involved in calculating this temperature rise is:\[ Q = m \, c \, \Delta T \]where:

• \( Q \) is the heat energy (in Joules),
• \( m \) is the mass of the objects involved (rock plus water, in kilograms),
• \( c \) is the specific heat capacity (a measure of how much energy it takes to change the temperature of a substance, different for each material),
• \( \Delta T \) is the change in temperature.

By rearranging the formula, you can solve for the temperature change (\( \Delta T \)) if you know the heat transferred (\( Q \)), the mass involved, and the specific heat capacity.

Specific heat capacity

Specific heat capacity is a property of a material that indicates how much energy is required to raise the temperature of 1 kilogram of that material by 1 degree Celsius. It is an essential concept in thermodynamics, helping us understand how different materials absorb or release heat.

Specific heat capacity is often represented by the symbol \( c \), and is measured in Joules per kilogram per degree Celsius (J/kg°C).
For example, the specific heat capacity of water is approximately \( 4186 \, \text{J/kg°C} \), which means it takes 4186 Joules of energy to raise the temperature of 1 kg of water by 1°C.

In the exercise, we calculated an average specific heat capacity (\( c_{eq} \)) for the rock and water system using the formula:\[ c_{eq} = \frac{m_{rock}c_{rock} + m_{water}c_{water}}{m_{total}} \]This helps account for the different materials interacting with the heat energy generated from the falling rock. The high specific heat capacity of water means it absorbs a lot of energy with only a small temperature change, which is why the temperature change calculated is relatively small.