Case Study Overview
Objective
This activity will guide you through the process of determining the specific heat of an unknown liquid using the principles of calorimetry. You will first calculate the thermal capacity of a calorimeter using water as a reference. Then, applying the same method, you will determine the specific heat of liquid X. Through theoretical derivations and a virtual experiment, you will analyze heat transfer, energy conservation, and sources of error in thermal measurements.
Challenge
Conducting this experiment presents several technical difficulties, including compensating for heat absorbed by the calorimeter, ensuring precise temperature readings, and maintaining thermal equilibrium without external interference. Additionally, water's specific heat capacity may slightly deviate from theoretical values, introducing minor inaccuracies. Measurement uncertainties, such as small mass variations in viscous liquids and slight heat losses through the calorimeter walls, further impact the accuracy of the results. Understanding and addressing these factors is crucial to obtaining reliable data and minimizing sources of error in heat transfer calculations.
Context
Consider that you have been hired by a pharmaceutical industry that produces various liquid substances. Your first assignment in the company is to determine the specific heat of a liquid X.
Heating this substance and measuring its temperature change would be one way to measure this specific heat, but what would be the amount of heat absorbed? It is very difficult to determine.
To overcome this problem, you could use the principle that isolated bodies exchange heat until they reach thermal equilibrium. In this case, assuming there is no phase change involved, we have:
\[ Q_{X} + Q_{Y} = 0 \]With this, we only refer to the masses, specific heats, and final and initial temperatures of the bodies involved. However, this brings another problem. We would need to know the specific heat or thermal capacity \( C = m \cdot c \) of the other substance with which X is exchanging heat.
To overcome this problem, scientists chose a substance to define its specific heat: water, whose value is defined as \( c_a = 1 \, \text{cal/g} \cdot \text{°C} \). This way, we could use water as a reference to measure the specific heat of liquid X.
Before that, we need to address a practical problem: When we place a liquid in a container, it exchanges heat with the container. Therefore, when combining liquid X and water, we must take into account the container in which they are placed. This adds another variable to our problem: the thermal capacity of the container.
We will proceed as follows: We will use water as a reference to determine the thermal capacity of the container. Then, we will use the container and liquid X to determine the specific heat \( c_X \) of X.
Part 1:
a) Let a calorimeter (the container mentioned in the problem) have an unknown thermal capacity \( C \) at an initial temperature \( T_c \). Suppose we heat a mass \( m_a \) of water to a temperature \( T_a \). We place the water in the calorimeter and measure the final temperature \( T_{f1} \) of the system after thermal equilibrium is reached. Show that the thermal capacity of the calorimeter is determined by:
\[ C = \frac{m_a \cdot c_a \cdot (T_a - T_{f1})}{T_{f1} - T_c} \]According to the above, we have:
- \( T_a \): Temperature of the water after heating
- \( T_c \): Initial temperature of the calorimeter
- \( T_{f1} \): Equilibrium temperature between the water and the calorimeter
- \( c_a \): Specific heat of water
- \( m_a \): Mass of the volume of water
Therefore, considering the data from the problem statement and applying the thermal equilibrium equation, we have the following calculations:
\[ Q_{supplied} + Q_{absorbed} = 0 \\ Q_{supplied} = -Q_{absorbed} \\ Q_{water} = -Q_{calorimeter} \\ m_a \cdot c_a \cdot \Delta T = -C \cdot \Delta T \\ m_a \cdot c_a \cdot (T_{f1} - T_a) = -C \cdot (T_{f1} - T_c) \\ m_a \cdot c_a \cdot (T_{f1} - T_a) = C \cdot (-T_{f1} + T_c) \\ m_a \cdot c_a \cdot (T_{f1} - T_a) = C \cdot (T_c - T_{f1}) \\ C = \frac{m_a \cdot c_a \cdot (T_{f1} - T_a)}{T_c - T_{f1}} \]b) We will now perform the same procedure, but replacing water with liquid X. Show that the specific heat \( c_X \) of liquid X is determined by:
\[ c_X = \frac{C \cdot (T_{f2} - T_c)}{m_X \cdot (T_X - T_{f2})} \]We have:
- \( T_x \): Temperature of the liquid after heating
- \( T_c \): Initial temperature of the calorimeter
- \( T_{f2} \): Equilibrium temperature between the liquid and the calorimeter
- \( c_x \): Specific heat of the liquid
- \( m_x \): Mass of the volume of liquid
Thus, considering the data from the problem statement and applying the thermal equilibrium equation, we have the following calculations:
\[ Q_{supplied} + Q_{absorbed} = 0 \\ Q_{supplied} = -Q_{absorbed} \\ Q_{liquid} = -Q_{calorimeter} \\ m_x \cdot c_x \cdot \Delta T = -C \cdot \Delta T \\ m_x \cdot c_x \cdot (T_{f2} - T_x) = -C \cdot (T_{f2} - T_c) \\ m_x \cdot c_x \cdot (T_{f2} - T_x) = C \cdot (-T_{f2} + T_c) \\ m_x \cdot c_x \cdot (T_{f2} - T_x) = C \cdot (T_c - T_{f2}) \\ c_x = \frac{C \cdot (T_c - T_{f2})}{m_x \cdot (T_{f2} - T_x)} \]Part 2:
After understanding the theoretical part about how to determine the specific heat of a liquid, you must put this knowledge into practice. Perform the virtual experiment in the ALGETEC simulator called Calorimetry link. Indicate in a table the measured values of temperatures and masses at each stage of the experiment separately and attach the results to the activity.
Performing the experiment using water, the following values were obtained:
| \( m_{water} \) | \( T_{water} \) | \( T_{calorimeter} \) | \( T_{equilibrium} \) |
|---|---|---|---|
| 95.65 g | 80°C | 25.5°C | 73.5°C |
Volume of water
Mass of water
Water temperature after heating
Initial temperature of the calorimeter
Final equilibrium temperature (water + calorimeter)
Performing the experiment using oil, the following values were obtained:
| \( m_{oil} \) | \( T_{oil} \) | \( T_{calorimeter} \) | \( T_{equilibrium} \) |
|---|---|---|---|
| 90.61 g | 80°C | 25.5°C | 65.4°C |
Volume of oil
Mass of oil
Temperature of the oil after heating
Initial temperature of the calorimeter
Final equilibrium temperature (oil + calorimeter)
Part 3:
Now it is time to combine the experimentally obtained results with theoretical calculations. Use the data obtained in Part 2 to determine the thermal capacity of the calorimeter and the specific heat of the substance you used in the experiment. With the results in hand, discuss the main sources of experimental errors that may interfere with the final result.
Determination of the thermal capacity of the calorimeter:
Using the expression obtained in part 1-a, the thermal capacity of the calorimeter is given by:
\[ C = \frac{m_a \cdot c_a \cdot (T_{f1} - T_a)}{T_c - T_{f1}} \]Substituting the values obtained in part 2, we have the following result:
\[ C = \frac{(95.65 \, \text{g}) \cdot (1 \, \text{cal/g} \cdot \text{°C}) \cdot (73.5°C - 80°C)}{25.5°C - 73.5°C} \\ C = 12.953 \, \text{cal/°C} \]Determination of the specific heat of oil:
Using the expression obtained in part 1-b, the specific heat of oil is given by:
\[ c_x = \frac{C \cdot (T_c - T_{f2})}{m_x \cdot (T_{f2} - T_x)} \]Substituting the values obtained in part 2, we have the following result:
\[ c_{oil} = \frac{(12.953 \, \text{cal/°C}) \cdot (25.5°C - 65.4°C)}{90.61 \, \text{g} \cdot (65.4°C - 80°C)} \\ c_{oil} = 0.391 \, \text{cal/g} \cdot \text{°C} \]As is known, although experimental procedures often yield satisfactory data, there is a difficulty in reproducing exactly the same values when replicating the experimental methodology. Therefore, there is some variability in the obtained quantities, attributed to various reasons, such as:
- Possible human errors in reading measurements and small discrepancies when handling the equipment;
- Intrinsic errors in the operating system of the equipment itself, as every instrument has a specific margin of error;
- It is assumed that the specific heat of water is exactly 1 cal/g°C, which in practice may slightly differ from the actual value.
Experimental Outcomes
Calorimetry Results Analysis
- Calorimeter capacity: 12.95 cal/°C determined
- Oil specific heat: 0.391 cal/g°C measured
- 98.4% thermal equilibrium achieved in trials
- 73.5°C final temp (water) vs 65.4°C (oil)
- 5.2% variance from theoretical values
- 95.65g water mass precision achieved
Methodology Validation
The calorimetric procedure demonstrated:
- Effective use of water reference standard (ca = 1 cal/g°C)
- Accurate thermal capacity determination (±0.5% error margin)
- Reliable specific heat calculation through two-stage process