Determination of Specific Heat Capacity of a Solid by Method of Mixtures
1. Aim
To determine the specific heat capacity of a given solid by the method of mixtures.
2. Apparatus Used
- Calorimeter with stirrer (made of copper)
- Calorimeter jacket (insulator)
- Laboratory thermometer (range: 0°C to 100°C)
- Sample of solid material (metal pieces like lead shots, copper, brass, etc.)
- Electronic balance
- Heating arrangement (Bunsen burner or water bath)
- Beaker (250 ml)
- Stop watch
- Thread
- Measuring cylinder (100 ml)
- Tongs
3. Diagram
4. Theory
The method of mixtures is based on the principle of conservation of energy. When two bodies at different temperatures are mixed, heat flows from the body at higher temperature to the body at lower temperature until thermal equilibrium is reached.
In this experiment, we heat a solid sample to a high temperature and then immerse it in water contained in a calorimeter at a lower temperature. The heat lost by the solid is equal to the heat gained by the water and the calorimeter. By measuring the initial and final temperatures, we can calculate the specific heat capacity of the solid.
According to the principle of calorimetry:
Heat lost by the hot body = Heat gained by the cold bodies
$$ m_s \times c_s \times (T_1 - T_3) = m_w \times c_w \times (T_3 - T_2) + m_c \times c_c \times (T_3 - T_2) $$
Where:
- $m_s$ = Mass of the solid sample
- $c_s$ = Specific heat capacity of the solid (to be determined)
- $T_1$ = Initial temperature of the solid
- $T_2$ = Initial temperature of water and calorimeter
- $T_3$ = Final temperature of the mixture at equilibrium
- $m_w$ = Mass of water
- $c_w$ = Specific heat capacity of water (4.18 J/g·°C)
- $m_c$ = Mass of calorimeter
- $c_c$ = Specific heat capacity of calorimeter material (for copper calorimeter, 0.38 J/g·°C)
5. Formula
From the energy conservation equation:
$$ m_s \times c_s \times (T_1 - T_3) = m_w \times c_w \times (T_3 - T_2) + m_c \times c_c \times (T_3 - T_2) $$
Solving for $c_s$:
$$ c_s = \frac{m_w \times c_w \times (T_3 - T_2) + m_c \times c_c \times (T_3 - T_2)}{m_s \times (T_1 - T_3)} $$
This can be simplified to:
$$ c_s = \frac{[m_w \times c_w + m_c \times c_c] \times (T_3 - T_2)}{m_s \times (T_1 - T_3)} $$
The term $[m_w \times c_w + m_c \times c_c]$ is also known as the water equivalent of the calorimeter.
6. Procedure
- Measure the mass of the empty calorimeter ($m_c$) using an electronic balance.
- Pour a measured quantity of water into the calorimeter and note its mass ($m_w$).
- Record the initial temperature ($T_2$) of the water and calorimeter using a thermometer.
- Measure the mass of the solid sample ($m_s$).
- Heat the solid sample in a water bath or using a Bunsen burner to a temperature ($T_1$) significantly above room temperature (typically 80-90°C). Keep it at this temperature for about 5 minutes to ensure uniform heating.
- Quickly transfer the heated solid to the calorimeter containing water. Cover the calorimeter with its lid and gently stir the contents with the stirrer.
- Record the temperature at regular intervals until it reaches a steady value ($T_3$), which is the final equilibrium temperature.
- Repeat the experiment at least three times for greater accuracy.
7. Observation Table
| Mass Measurements | Temperature Measurements | |||
|---|---|---|---|---|
| Mass of calorimeter ($m_c$) | _____ g | Initial temperature of solid ($T_1$) | _____ °C | Temperature vs Time Graph |
| Mass of water ($m_w$) | _____ g | Initial temperature of water and calorimeter ($T_2$) | _____ °C | |
| Mass of solid ($m_s$) | _____ g | Final equilibrium temperature ($T_3$) | _____ °C | |
Temperature Readings vs Time
| Time (seconds) | Temperature (°C) | Time (seconds) | Temperature (°C) |
|---|---|---|---|
| 0 | 150 | ||
| 30 | 180 | ||
| 60 | 210 | ||
| 90 | 240 | ||
| 120 | 270 |
8. Calculations
Given data from observation table:
- Mass of calorimeter ($m_c$) = _____ g
- Mass of water ($m_w$) = _____ g
- Mass of solid ($m_s$) = _____ g
- Initial temperature of solid ($T_1$) = _____ °C
- Initial temperature of water and calorimeter ($T_2$) = _____ °C
- Final equilibrium temperature ($T_3$) = _____ °C
- Specific heat capacity of water ($c_w$) = 4.18 J/g·°C
- Specific heat capacity of calorimeter (copper) ($c_c$) = 0.38 J/g·°C
Using the formula:
$$ c_s = \frac{[m_w \times c_w + m_c \times c_c] \times (T_3 - T_2)}{m_s \times (T_1 - T_3)} $$
Substituting the values:
$$ c_s = \frac{[_____ \times 4.18 + _____ \times 0.38] \times (_____ - _____)}{_____ \times (_____ - _____)} $$
$$ c_s = _____ \text{ J/g·°C} $$
9. Result
The specific heat capacity of the given solid is _____ J/g·°C or _____ J/kg·K.
Comparing with standard values:
| Material | Standard Value (J/g·°C) | Experimental Value (J/g·°C) | Percentage Error (%) |
|---|---|---|---|
| _______ (name of solid) | _______ | _______ | _______ |
Percentage Error calculation:
$$ \text{Percentage Error} = \left| \frac{\text{Standard Value} - \text{Experimental Value}}{\text{Standard Value}} \right| \times 100\% $$
10. Precautions
- The calorimeter should be well-insulated to minimize heat exchange with the surroundings.
- The temperature of the solid should be measured just before transferring it to the calorimeter.
- The transfer of the solid to the calorimeter should be done quickly to minimize heat loss.
- Stir the water gently to ensure uniform temperature distribution.
- Avoid direct contact between the thermometer and the solid.
- Ensure that the calorimeter is completely dry before starting the experiment.
- Record all measurements with appropriate precision.
- Care should be taken while handling hot objects.
- The initial temperature of water should be 2-3°C below room temperature to account for heat gain from surroundings.
- Ensure the solid is completely immersed in water during the experiment.
11. Sources of Error
- Heat Loss: Heat exchange with the surroundings despite using an insulated calorimeter.
- Radiation Losses: Heat loss due to radiation while transferring the hot solid to the calorimeter.
- Thermometer Errors: Inaccuracies in thermometer readings or slow response time.
- Impurities: Presence of impurities in the solid sample or water.
- Evaporation: Loss of water due to evaporation during the experiment.
- Heat Capacity of Thermometer: Neglecting the heat absorbed by the thermometer.
- Non-uniform Heating: The solid may not be uniformly heated throughout its mass.
- Reading Errors: Human errors in reading measurements.
- Heat Transfer Delay: Time lag in heat transfer between the solid and water.
- Calibration Errors: Inaccuracies in the calibration of measuring instruments.
12. Viva Voice Questions
Q1: What is specific heat capacity?
A1: Specific heat capacity is the amount of heat energy required to raise the temperature of 1 gram of a substance by 1 degree Celsius. It is usually denoted by 'c' and measured in J/g·°C or J/kg·K.
Q2: Why is water used as a reference substance in calorimetry?
A2: Water has a high specific heat capacity (4.18 J/g·°C), which makes it an excellent heat reservoir. It helps in achieving measurable temperature changes when mixed with other substances. Moreover, water is readily available, inexpensive, and has well-established thermal properties.
Q3: What is the principle behind the method of mixtures?
A3: The method of mixtures is based on the principle of conservation of energy. It states that in an isolated system, the total heat lost by hot bodies equals the total heat gained by cold bodies until thermal equilibrium is reached.
Q4: Why do we use a calorimeter made of copper?
A4: Copper has high thermal conductivity, which ensures rapid and uniform distribution of heat throughout the calorimeter. It also has a relatively low specific heat capacity, reducing the error in calculations due to heat absorbed by the calorimeter itself.
Q5: What is water equivalent?
A5: Water equivalent is the mass of water that would absorb or release the same amount of heat as the calorimeter for the same temperature change. Mathematically, it is represented as $m_w + (m_c \times \frac{c_c}{c_w})$.
Q6: How does the specific heat capacity vary with temperature?
A6: For most substances, specific heat capacity slightly increases with increasing temperature. However, for practical purposes, it is often considered constant over moderate temperature ranges.
Q7: Why is stirring necessary during the experiment?
A7: Stirring ensures uniform temperature distribution throughout the mixture, preventing temperature gradients and ensuring that the final equilibrium temperature is accurate.
Q8: How would the results be affected if some water splashed out when the solid was dropped in?
A8: If water splashes out, the actual mass of water involved in the heat exchange would be less than the recorded value. This would lead to an overestimation of the specific heat capacity of the solid.
Q9: Why should the initial temperature of water be slightly below room temperature?
A9: If the initial temperature of water is slightly below room temperature, the heat gained from the surroundings during the experiment will help to compensate for the heat lost to the surroundings when the hot solid is added, minimizing the net external heat exchange.
Q10: How would you modify this experiment to determine the specific heat capacity of a liquid?
A10: To determine the specific heat capacity of a liquid, we would need to use a different method, such as the continuous flow method or the electrical heating method. In a modified method of mixtures, we could heat the liquid in a sealed container (to prevent evaporation) and then immerse it in water in a calorimeter, measuring the heat exchange.