Problem 91 - Physics

Question

A rock of mass \(0.20 \mathrm{~kg}\) 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 rise in temperature is approximately 0.015°C.

Step 1: Calculate Potential Energy

First, determine the potential energy (PE) of the rock when it is at a height of 15 meters. Use the formula: \( PE = m \cdot g \cdot h \), where \( m = 0.20 \text{ kg} \), \( g = 9.81 \text{ m/s}^2 \), and \( h = 15 \text{ m} \). Thus, \( PE = 0.20 \times 9.81 \times 15 \).

Step 2: Compute Potential Energy

Calculate the result from the previous step: \( PE = 0.20 \times 9.81 \times 15 = 29.43 \text{ J} \). This is the energy converted to heat when the rock hits the water.

Step 3: Calculate Total Mass and Specific Heat Capacity

Add the mass of the rock and water to get the total mass: \( 0.20 + 0.35 = 0.55 \text{ kg} \). Next, find the weighted average specific heat capacity: \( c_{avg} = \frac{m_{rock} \cdot c_{rock} + m_{water} \cdot c_{water}}{m_{total}} \). With \( c_{rock} = 1840 \text{ J/kg}^\circ\text{C} \) and \( c_{water} = 4184 \text{ J/kg}^\circ\text{C} \), \( c_{avg} \approx 3515.27 \text{ J/kg}^\circ\text{C} \).

Step 4: Calculate Temperature Rise

Use the formula for temperature change: \( \Delta T = \frac{PE}{m_{total} \cdot c_{avg}} \). Substituting the known values: \( \Delta T = \frac{29.43 \text{ J}}{0.55 \text{ kg} \times 3515.27 \text{ J/kg}^\circ\text{C}} \).

Step 5: Compute Temperature Increase

Calculate the temperature rise: \( \Delta T = \frac{29.43}{1933.40} \approx 0.015 ^\circ\text{C} \).

Key Concepts

Potential Energy

Potential energy is the energy stored in an object due to its position relative to a reference point. In this case, the rock situated at a height of 15 meters has gravitational potential energy due to the force of gravity acting on it. The potential energy can be calculated using the formula:

• \( PE = m \cdot g \cdot h \)

Here:

• \(m\) is the mass of the rock, which is \(0.20 \text{ kg}\).
• \(g\) is the acceleration due to gravity, commonly taken as \(9.81 \text{ m/s}^2\).
• \(h\) is the height from which the rock falls, which is \(15 \text{ m}\).

Plugging in these values, the potential energy stored in the rock before it falls is \(29.43 \text{ J}\). This energy is transformed into kinetic energy as the rock falls and upon reaching the ground, it turns into heat energy, warming up the water.

Specific Heat Capacity

Specific heat capacity is a crucial concept in thermodynamics that describes the amount of heat required to change the temperature of a substance by one degree Celsius. It is expressed in units of \(\text{J/kg} \cdot\text{°C}\). A higher specific heat capacity indicates that the substance requires more heat to increase its temperature, implying it can store more energy.

In the given problem:

• The specific heat capacity of the rock is given as \(1840 \text{ J}/(\text{kg}\cdot \text{°C})\), meaning that \(1840 \text{ J}\) of energy is required to raise one kilogram of the rock by one °C.
• Water, conversely, has a specific heat capacity of \(4184 \text{ J}/(\text{kg}\cdot \text{°C})\), which is over twice that of the rock.

These values indicate how energy from the rock's fall is distributed between the rock and the water. The final step involves averaging the specific heat capacities, weighted by their masses, to find how this energy affects the entire system.

Temperature Change

Temperature change (\(\Delta T\)) is an important factor we calculate to understand how much the temperature of an object has altered due to an energy exchange or conversion. In this scenario, after the energy from the potential energy is converted, it is crucial to determine by how much this energy changes the temperature inside the pail.

This can be calculated using the formula:

• \( \Delta T = \frac{PE}{m_{\text{total}} \cdot c_{\text{avg}}} \)

Where:

• \(PE\) is the potential energy converted into heat (\(29.43 \text{ J}\)).
• \(m_{\text{total}}\) is the total mass of the rock and water (\(0.55 \text{ kg}\)).
• \(c_{\text{avg}}\) is the weighted average specific heat capacity of the system.

Using these values allows us to find that the temperature change experienced by the rock and water in the pail is very small, only increasing by approximately \(0.015 \text{°C}\).

Energy Conversion

Understanding energy conversion helps explain how energy changes forms during different processes. Here, energy conversion is illustrated through the transforming of potential energy (due to height) into heat, which is then absorbed by the water and rock, causing a temperature increase.

This conversion process highlights a few key principles:

• **Conservation of Energy**: The total energy of an isolated system remains constant, merely converting from one form to another.
• **Efficiency**: Not all potential energy becomes useful energy (for temperature change); some are lost to the surroundings or absorbed otherwise.

For students grappling with thermodynamics, this exercise underscores how energy forms like gravitational potential energy, kinetic energy, and thermal energy are interconnected and the physical laws governing their transformation.