LoginGet started

Thermal Physics | Practical No 26B - Specific heat capacity of liquid | A/L Physics Practical

DP Education - A/L සිංහල මාධ්‍යය・2 minutes read

The experiment aims to determine the specific heat capacity of an unknown liquid by measuring the cooling rates of both water and the liquid in a controlled environment, utilizing a calorimeter and continuous airflow to maintain consistent conditions. The specific heat capacity is calculated by equating the heat loss rates of the two liquids, taking into account their mass and temperature changes, with careful attention to the materials and setup to ensure accurate results.

Insights

  • The experiment seeks to find the specific heat capacity of an unknown liquid by measuring heat loss rates from two calorimeters, one containing water and the other containing the unknown liquid, while maintaining consistent conditions such as surface area and surrounding temperature to ensure valid comparisons.
  • The heat loss rate is influenced by factors like the material of the calorimeter, the surface area, and the temperature difference between the liquid and its surroundings, with the formula \( \text{Heat loss rate} = k \cdot A \cdot (\theta_2 - \theta_1) \) being key to understanding how these variables interact.
  • Continuous stirring of the liquids and the use of an electric fan are essential to maintain uniform temperature distribution and consistent cooling conditions, preventing localized temperature variations that could skew the results of the specific heat capacity calculations.
  • The results of the experiment show that water cools more slowly than the unknown liquid, with specific heat capacities calculated using the formula \( c = \frac{Q}{m \cdot \Delta T} \), demonstrating the importance of accurate measurements and the effects of different materials on heat retention and loss.

Get key ideas from YouTube videos. It’s free

Recent questions

  • What is specific heat capacity?

    Specific heat capacity is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). It is a crucial property in thermodynamics, as it indicates how much energy a material can store and how it responds to heat changes. Different substances have varying specific heat capacities, which affect their heating and cooling rates. For example, water has a high specific heat capacity, meaning it can absorb a lot of heat without a significant temperature change, making it effective for temperature regulation in natural and engineered systems. Understanding specific heat capacity is essential in fields such as chemistry, physics, and engineering, as it influences processes like heating, cooling, and energy transfer.

  • How to measure heat loss?

    Measuring heat loss involves determining the amount of thermal energy that escapes from a substance or system over time. This can be done using calorimetry, where a calorimeter is used to contain the substance while monitoring temperature changes. The heat loss rate can be calculated using the formula \( \text{Heat loss rate} = k \cdot A \cdot (\theta_2 - \theta_1) \), where \( k \) is a constant related to the material's surface, \( A \) is the surface area, and \( \theta_2 - \theta_1 \) is the temperature difference between the substance and its surroundings. By recording temperature changes at regular intervals and knowing the mass and specific heat capacity of the substance, one can accurately quantify the heat lost during the cooling process. This method is widely used in experiments to study thermal properties and energy efficiency.

  • What is Newton's law of cooling?

    Newton's law of cooling states that the rate of heat loss of an object is directly proportional to the difference in temperature between the object and its surrounding environment, provided that this temperature difference is small. This principle is fundamental in thermodynamics and is often expressed mathematically as \( \frac{dT}{dt} = -k(T - T_{env}) \), where \( T \) is the temperature of the object, \( T_{env} \) is the ambient temperature, and \( k \) is a constant that depends on the characteristics of the object and the environment. This law helps predict how quickly an object will cool down or heat up, which is essential in various applications, including engineering, environmental science, and food safety. Understanding this law allows for better control of temperature in processes where heat transfer is critical.

  • Why use a calorimeter?

    A calorimeter is used to measure the heat transfer associated with chemical reactions or physical changes, such as temperature changes in a substance. It provides a controlled environment to accurately assess the specific heat capacity of materials by minimizing heat exchange with the surroundings. The design of a calorimeter, often insulated and made from materials with known thermal properties, allows for precise measurements of temperature changes over time. By using a calorimeter, researchers can determine the amount of heat absorbed or released during a process, which is crucial for understanding thermodynamic properties and behaviors of substances. This makes calorimeters essential tools in chemistry, physics, and engineering for studying energy changes in various systems.

  • What factors affect cooling rates?

    Cooling rates are influenced by several factors, including the temperature difference between the object and its environment, the surface area of the object, the nature of the surface material, and the surrounding airflow. According to Newton's law of cooling, a greater temperature difference results in a faster cooling rate. Additionally, larger surface areas allow for more heat to escape, enhancing the cooling process. The material's thermal conductivity also plays a significant role; for instance, metals typically cool faster than insulators due to their ability to transfer heat efficiently. Lastly, external conditions such as air movement can affect cooling rates; for example, using a fan to increase airflow around a cooling object can significantly enhance heat loss. Understanding these factors is essential for accurately predicting and controlling cooling processes in various applications.

Related videos

Summary

00:00

Determining Specific Heat of Unknown Liquid

  • The experiment aims to determine the specific heat capacity of an unknown liquid using the method of cooling, focusing on the heat loss from two warm bodies in a continuous air stream.
  • The heat loss rate formula is given as \( \text{Heat loss rate} = k \cdot A \cdot (\theta_2 - \theta_1) \), where \( k \) is a constant based on the surface material, \( A \) is the surface area, and \( \theta_2 - \theta_1 \) is the excess temperature difference between the surface and the surroundings.
  • The experiment uses a calorimeter filled with hot water at 80°C and surrounding air at 30°C, establishing that heat will flow from the hotter water to the cooler environment.
  • The constant \( k \) varies with the material of the calorimeter (e.g., copper vs. aluminum), and the heat loss rate increases with larger surface areas and greater temperature differences.
  • Two identical calorimeters are used, one filled with water and the other with an unknown liquid, ensuring that \( k \), \( A \), and the surrounding temperature are the same for both setups.
  • The cooling process is maintained by using an electric fan to ensure a continuous flow of air, preventing the surrounding air temperature from rising above 30°C.
  • The heat loss rate can also be expressed as \( \frac{m \cdot c \cdot \Delta \theta}{\Delta t} \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta \theta \) is the change in temperature over time \( \Delta t \).
  • The specific heat capacity of the unknown liquid is determined by equating the heat loss rates of both the water and the unknown liquid, using measured values for mass and temperature changes.
  • The experimental setup requires specific materials: a polished calorimeter with a lid and stirrer, a thermometer, an electric fan, a stopwatch, a triple beam balance, and sufficient quantities of water and the unknown liquid.
  • The procedure involves measuring the mass of the calorimeter, filling it with water to 1 cm below the top, heating it to 80°C, and ensuring uniform temperature distribution to accurately measure the cooling rates.

20:43

Comparing Cooling Rates of Water and Oil

  • Fill the calorimeter with water, ensuring it is one centimeter below the top to prevent spillage during stirring, which is essential for maintaining an even temperature throughout the water and the calorimeter walls.
  • Stir the water continuously throughout the experiment to ensure uniform temperature distribution, as stagnant water may have varying temperatures, affecting the accuracy of measurements.
  • Use an electric fan to cool the calorimeter by maintaining a constant airflow while stirring, and record the temperature of the water every 30 seconds until it cools down to approximately 40 degrees Celsius.
  • Start measuring the temperature at 80 degrees Celsius and continue recording every 30 seconds until reaching 40 degrees, noting down all temperature readings for later graphing.
  • Measure the mass of the calorimeter filled with water to determine the mass of the water by subtracting the mass of the empty calorimeter from the total mass.
  • After completing the water experiment, empty and clean the calorimeter, then fill it with an equal volume of heated coconut oil, ensuring the oil level is the same as the water level during the previous experiment.
  • Heat the coconut oil to 80 degrees Celsius, ensuring it does not exceed this temperature to avoid flammability, and then start the stopwatch to record the temperature every 30 seconds as it cools down to 40 degrees.
  • Use a three-beam balance to measure the mass of the calorimeter with the oil, allowing for the calculation of the mass of the oil by subtracting the mass of the empty calorimeter.
  • Draw cooling curves for both water and coconut oil on the same graph, plotting temperature on the y-axis and time on the x-axis, to compare the cooling rates and calculate the specific heat capacities.
  • Equate the heat loss rates for water and coconut oil using the formula \( m c \Delta \theta / \Delta t \) to find the specific heat capacity of the coconut oil, ensuring that the surface area and conditions remain constant throughout the experiment.

45:14

Calorimeter Experiment for Heat Capacity Analysis

  • To maintain constant conditions during the experiment, use thermal insulators; three pieces of caulk are recommended to be placed at the bottom of a beaker to prevent heat exchange with the environment.
  • Place a calorimeter with weights on top of the thermal insulators, ensuring it does not float, and fill the gap between the beaker and calorimeter with cool water to the height of the calorimeter.
  • The calorimeter will contain either hot water or oil, starting at a temperature of 80 degrees Celsius, and will be allowed to cool down to 40 degrees Celsius while measuring the temperature at regular intervals.
  • Suspend the calorimeter using a string so that it does not touch the other calorimeter, allowing for accurate temperature readings without interference.
  • Use cool water in the outer beaker due to its high specific heat capacity, which allows it to absorb heat rapidly, ensuring the surrounding temperature remains constant during the experiment.
  • If switching from water to oil, replace the outer water with cool water again to maintain consistent conditions, as any heat gained by the outer water could affect the results.
  • Record temperature readings every 30 seconds, noting that the mass of the calorimeter and water is 408.8 grams, while the mass of the oil used is 384 grams.
  • For graphing the cooling curves, use a graph sheet with time in minutes on the x-axis and temperature in degrees Celsius on the y-axis, marking the maximum temperature at 80 degrees and the minimum at 40 degrees.
  • The cooling curve for oil shows a rapid initial drop in temperature that slows over time, while water, due to its higher specific heat capacity, cools more slowly, taking approximately 2.5 minutes to cool from 60 to 50 degrees Celsius compared to 1.22 minutes for oil.
  • Perform calculations using the formula \( m c \Delta T / \Delta t \) for both water and oil, ensuring to convert all measurements to consistent units, and use the recorded values to derive specific heat capacities and validate the experiment's findings.

01:13:13

Determining Specific Heat Capacity of Liquids

  • The process of determining the specific heat capacity of a liquid involves using a calorimeter filled with the liquid, which is placed on coke pieces within a white beaker, ensuring that the cooling conditions are consistent for both the liquid and water used in the experiment.
  • To calculate the specific heat capacity, one must draw cooling curves for both the water and the unknown liquid on the same graph, measuring the time taken for each to cool from one temperature to another, while ensuring equal quantities of both liquids are used to maintain consistent surface area and cooling conditions.
  • The mean rate of heat loss from the calorimeter and the liquid can be expressed mathematically, where the heat capacity of the calorimeter and stirrer is denoted as \(C\), the mass of water as \(m_w\), the mass of the liquid as \(m_l\), and their respective specific heat capacities as \(s_w\) and \(s_l\).
  • Newton's law of cooling states that the cooling rate is proportional to the excess temperature between the object and its surroundings, and this relationship holds true only if the surrounding conditions remain constant, such as maintaining constant airflow.
  • The experiment requires careful attention to units, as all calculations must include the correct units for energy, mass, and temperature, ensuring that the final answer for specific heat capacity is expressed in joules per kilogram per kelvin (J/kg·K).
  • A metallic calorimeter, preferably made of copper or aluminum, is essential for accurate results, as glass beakers cannot be used due to their poor heat conduction, which would lead to temperature discrepancies between the water and the beaker's surface.
  • When conducting the experiment, it is crucial to fill the calorimeter to a level that ensures the entire surface is maintained at a uniform temperature, ideally just below the top, to avoid temperature gradients that could affect the cooling rate.
  • Stirring the liquid continuously during the experiment is necessary to ensure that the thermometer accurately reflects the temperature of the liquid and the calorimeter surface, preventing localized temperature variations.
  • The experiment's design must account for the heat loss rate, which is influenced by factors such as surface area, the nature of the surface, and the excess temperature, all of which must be controlled to ensure valid results.
  • Past exam questions emphasize the importance of understanding the principles behind each step of the experiment, as well as the need to practice with previous papers to familiarize oneself with the types of questions that may arise regarding the specific heat capacity determination process.

01:37:26

Heat Loss Experiment Reveals Measurement Challenges

  • The experiment involves measuring the heat loss of water and a liquid, where water's temperature changes from 55°C to 45°C over 4 minutes, resulting in a heat loss rate of 38 joules per second. The specific heat capacity of the liquid is calculated using the formula \( c = \frac{Q}{m \cdot \Delta T} \), where \( Q \) is the heat loss (38 joules), \( m \) is the mass of the liquid (0.172 kg), and \( \Delta T \) is the temperature difference (10°C), leading to a specific heat capacity of 2000 joules per kilogram per kelvin.
  • The experiment highlights the unsuitability of using a glass container instead of a calorimeter, as the glass surface temperature does not match the water's temperature, leading to inaccurate readings. The heat loss occurs from the glass surface, which can differ from the water temperature, affecting the experiment's accuracy.
Channel avatarChannel avatarChannel avatarChannel avatarChannel avatar

Try it yourself — It’s free.