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Year 11 Chemistry Practical Investigation | Calorimetry Experiment

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How to perform the calorimetry experiment in Year 11 Chemistry Practical

A popular Year 11 Chemistry practical investigation is the calorimetry experiment. Not only is this experiment commonly performed by students during their Year 11 Chemistry course but also in the HSC Chemistry course. In this article, you will find a complete Chemistry practical report on determining the enthalpy of combustion of fuels via calorimetry.

This Year 11 Chemistry practical report on the calorimetry experiment consists of:

• Safety information 
• Practice calorimetry problems

Calorimetry Experiment

To determine the enthalpy of combustion of fuels using a calorimeter.

The standard enthalpy of combustion \small (\Delta H_c^\circ) is the enthalpy change when one mole of a substance undergoes complete combustion with oxygen at standard states, under standard conditions.

The steps which can be used to determine the enthalpy changes of combustion are outlined below:

Step 1: Write a balanced chemical equation of the process.

\large \text{Fuel}_{(s)/(l)/(g)} + {O_2}_{(g)} \rarr {CO_2}_{(g)} + {H_2O}_{(l)}

Step 2: Calculate the heat gained by the substance (water).

\Large q_\text{ substance} =mc\Delta T

• \small q_\text{ substance} is the heat gained by water in joules \small \text{(J)}
• \small \text{m} is the mass of water in kilograms \small (\text{kg})
• \small \text{c} is the specific heat capacity of water (4.18 × 10 3 \small \text{J kg}^{-1}\text{K}^{-1}   or 4.18 \small \text{J g}^{-1}\text{K}^{-1} )
• \small \Delta \text{ T} is the change in temperature of water in kelvin \small \text{(K)}

Step 3: Calculate the heat released by the combustion process.

The quantity of heat exchanged between the process and the substance will be the same but opposite sign .

\Large q_\text{ combustion process} = -q_\text{ substance}

Step 4: Calculate the enthalpy change of the process.

To calculate the standard enthalpy of combustion from the results of a calorimetry experiment:

\Large \Delta H_c = \dfrac{q_\text{ combustion process}}{n_\text{ combustion process}}

Since the standard enthalpy of combustion of fuels (∆H c ) is always negative, we often use the term “heat of combustion”. The heat of combustion is the absolute value of the standard enthalpy of combustion, as the amount of heat released for a specified amount of the fuel.

The equation q=mc\Delta T and the value for the specific heat capacity of water can be found on the HSC Chemistry Formula and Data Sheet .

The fuels used during this experiment are alcohols which have a general formula:

\Large C_nH_{2n+1}OH .

For example,

• Methanol has the formula \small CH_3OH .
• Ethanol has the formula \small C_2H_5OH .
• Propanol has the formula \small C_3H_7OH .
• Retort stand and clamp
• Methanol spirit burner
• Ethanol spirit burner
• Propan-1-ol spirit burner
• 250 \small \text{mL} measuring cylinder
• Thermometer (0 – 100 \small \degree \text{C} accurate to 0.1 \small \degree \text{C} )
• Electronic balance

Safety Information

• Set up the equipment as shown below. The copper can should be clamped so that the tip of the flame just touches the can when lit.
• Add 200 \small \text{mL} of cold water to the can using a measuring cylinder to measure the volume of water.
• Record the initial temperature of the water using a thermometer.
• Weigh the spirit burner with its liquid contents as accurately as possible, and record the mass.
• Light the wick and stir the water gently with a glass rod. Monitor the temperature and observe the flame.
• When the temperature has risen by about 10 \small \degree \text{C} , extinguish the flame by replacing the cap.
• Accurately record the maximum temperature for the water.
• Reweigh the burner and record its final mass.
• Examine the bottom of the can for soot accumulation. Remove soot before using the next alcohol.

Table 1: Experimental measurements and observations

Calorimetry formulas

Calorimetry calculations.

An example of the calculation of the enthalpy of combustion of methanol \small (CH_3OH) is shown below.

{CH_3OH}_{(l)} + \frac{3}{2} {O_2}_{(g)} \rarr 2{H_2O}_{(l)} + {CO_2}_{(g)}

Step 2: Calculate the heat gained by the water.

\begin{aligned} q_\text{ substance} & = mc\Delta T \\ \\ & = 200 \text{ kg} \times (4.18 \times 10^{-3} \text{J kg}^{-1}\text{K}^{-1}) \times (35.0-25.0) \text{ K} \\ \\ &= 8360 \text{ J} \end{aligned}

\begin{aligned} q_\text{ combustion process} &= -q_\text{ substance} \\ \\ &=-8360 \text{ J} \\ \\ &=-8.360 \text{ kJ} \end{aligned}

\begin{aligned} n(CH_3OH) & = \dfrac{n}{MM} \\ \\ &=\dfrac{0.55}{12.01+4\times 1.008+16} \\ \\ &=0.01716 \dots \text{ mol} \end{aligned}

\begin{aligned} \Delta H_c&=\dfrac{q_\text{ combustion process}}{n_\text{ combustion process}} \\ \\ & = -\dfrac{8.36}{0.01716 \dots} \\ \\ & = - 490 \text{ kJ mol}^{-1} \text{(2 s.f.)} \end{aligned}

A summary of the calculations for the enthalpy of combustion of methanol, ethanol and propan-1-ol is displayed in the table below.

Table 2: Calculations

Students are often asked to answer the following quantitative and qualitative analysis questions after performing a chemistry practical on calorimetry.

1. The theoretical value for the enthalpy of combustion of each alcohol is given in the table below. 

Calculate the percentage error of the experimental result compared to the theoretical result for each alcohol. 

\% \text{ error} = \dfrac{| \text{theoretical value} - \text{experimental value}|}{\text{theoretical value}} \times 100

\begin{aligned} \% \text{ error}_\text{ methanol} &= \dfrac{|-726 -(-490)|}{726} \times 100 \\ \\ & = 32.5 \% \\ \\ \% \text{ error}_\text{ ethanol} &= \dfrac{|-1368 -(-850)|}{1368} \times 100 \\ \\ &=37.9\%\\ \\ \% \text{ error}_\text{ propan-1-ol} &= \dfrac{|-2021 -(-1200)|}{2021} \times 100 \\ \\ &=40.6 \% \end{aligned}

Let’s investigate the safety, errors, reliability and accuracy of this experiment.

1. Outline two safety risks in this experiment. Describe how the risks were minimised. 

2. Account for the differences between the experimental value and the theoretical values for the standard enthalpies of combustion. 

In the experiment, not all of the heat produced was used in heating the water. Some of the heat was lost to the surroundings (heating the copper can and the air around the flame) . Since this was not taken into account in the calculations, it caused the experimental value for the enthalpy of combustion to be significantly higher than the theoretical value.

3. Assess the validity of this experiment

Validity relates to the experimental method and how appropriate it is in addressing the aim of the experiment.

The aim of this experiment was to determine the enthalpy of combustion of fuels using a calorimeter. Therefore, the validity of the experiment can be assessed based on how suitable the method was in determining the enthalpy of combustion of each fuel. 

• The enthalpy of combustion of a fuel refers to the energy released in the complete combustion of one mole of fuel. Therefore the fuels should have undergone complete combustion. However, this is not true for ethanol and propan-1-ol as black soot was observed on the base of the copper can. Black soot which is C (s) is an indication of incomplete combustion. The blue-yellow and orange flame observed during the combustion of ethanol and propan-1-ol is also an indication of incomplete combustion.
• The calculation of enthalpy made in this experiment assumes that there is no heat loss. However, this assumption is not satisfied as considerable heat is lost to the surroundings. 

Since the experimental method contains assumptions that are not valid, the experiment is not valid.

4. Suggest techniques that could be used to improve the validity of the results.

The validity of an experiment can be improved by:

• Keeping the control variables constant and preventing them from affecting the dependent variable.
• Ensuring that any assumptions made are valid.

5. What can be done to ensure the reliability of the results?

Reliability is the extent to which the experiment yields the same result each time.

The reliability can be improved by conducting more trials for each fuel, excluding outliers and averaging concordant results. Using a greater volume of water and recording results over a larger change in temperature will also reduce the percentage error, minimise the effect of random errors and improve the reliability of the results.

6. What can be done to improve the accuracy of the results?

Accuracy is the extent to which the calculated value differs from the true, accepted value.

Eliminating systematic errors arising from the incorrect use of equipment or improper calibration of instruments such as zero setting error (where the instrument does not read zero when the quantity to be measured is zero) will improve accuracy.

Using more precise measuring devices will also improve accuracy. For example, use a 0–50 \small \degree \text{C} thermometer. It will be more accurate than a 0–100 \small \degree \text{C} one as the scale divisions are smaller. Alternatively, use a digital thermometer.

For more information on how to perform quantitative and qualitative data analysis on chemistry practical investigations, read the guide on How to study on data analysis task

A calorimetry experiment was conducted to determine the molar enthalpy of combustion of ethanol \small (C_2H_5OH) , molar mass = 46.07 \small \text{g mol}^{-1} ). The following data were collected:

Question 2 

One method for improving the experimental design is to take into account the energy absorbed by the calorimeter. This energy is then added to the energy absorbed by the water to calculate the enthalpy of combustion.

Use the measurements provided below to calculate the approximate enthalpy of combustion of butan-1-ol ( \small MM =74.12 \small \text{g mol}^{-1} ). (4 marks)

Solutions to calorimetry problems

Written by Varisara Laosuksri

Varisara is a 2019 St George Girls High School graduate who achieved Band 6 in her HSC Chemistry and Physics.

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1.4 Heat Transfer, Specific Heat, and Calorimetry

Learning objectives.

By the end of this section, you will be able to:

• Explain phenomena involving heat as a form of energy transfer
• Solve problems involving heat transfer

We have seen in previous chapters that energy is one of the fundamental concepts of physics. Heat is a type of energy transfer that is caused by a temperature difference, and it can change the temperature of an object. As we learned earlier in this chapter, heat transfer is the movement of energy from one place or material to another as a result of a difference in temperature. Heat transfer is fundamental to such everyday activities as home heating and cooking, as well as many industrial processes. It also forms a basis for the topics in the remainder of this chapter.

We also introduce the concept of internal energy, which can be increased or decreased by heat transfer. We discuss another way to change the internal energy of a system, namely doing work on it. Thus, we are beginning the study of the relationship of heat and work, which is the basis of engines and refrigerators and the central topic (and origin of the name) of thermodynamics.

Internal Energy and Heat

A thermal system has internal energy , which is the sum of the microscopic energies of the system. This includes thermal energy, which is associated with the mechanical energies of its molecules and which is proportional to the system’s temperature. As we saw earlier in this chapter, if two objects at different temperatures are brought into contact with each other, energy is transferred from the hotter to the colder object until the bodies reach thermal equilibrium (that is, they are at the same temperature). No work is done by either object because no force acts through a distance (as we discussed in Work and Kinetic Energy ). These observations reveal that heat is energy transferred spontaneously due to a temperature difference. Figure 1.9 shows an example of heat transfer.

The meaning of “heat” in physics is different from its ordinary meaning. For example, in conversation, we may say “the heat was unbearable,” but in physics, we would say that the temperature was high. Heat is a form of energy flow, whereas temperature is not. Incidentally, humans are sensitive to heat flow rather than to temperature.

Since heat is a form of energy, its SI unit is the joule (J). Another common unit of energy often used for heat is the calorie (cal), defined as the energy needed to change the temperature of 1.00 g of water by 1.00 ° C 1.00 ° C —specifically, between 14.5 ° C 14.5 ° C and 15.5 ° C 15.5 ° C , since there is a slight temperature dependence. Also commonly used is the kilocalorie (kcal), which is the energy needed to change the temperature of 1.00 kg of water by 1.00 ° C 1.00 ° C . Since mass is most often specified in kilograms, the kilocalorie is convenient. Confusingly, food calories (sometimes called “big calories,” abbreviated Cal) are actually kilocalories, a fact not easily determined from package labeling.

Mechanical Equivalent of Heat

It is also possible to change the temperature of a substance by doing work, which transfers energy into or out of a system. This realization helped establish that heat is a form of energy. James Prescott Joule (1818–1889) performed many experiments to establish the mechanical equivalent of heat — the work needed to produce the same effects as heat transfer . In the units used for these two quantities, the value for this equivalence is

We consider this equation to represent the conversion between two units of energy. (Other numbers that you may see refer to calories defined for temperature ranges other than 14.5 ° C 14.5 ° C to 15.5 ° C 15.5 ° C .)

Figure 1.10 shows one of Joule’s most famous experimental setups for demonstrating that work and heat can produce the same effects and measuring the mechanical equivalent of heat. It helped establish the principle of conservation of energy. Gravitational potential energy ( U ) was converted into kinetic energy ( K ), and then randomized by viscosity and turbulence into increased average kinetic energy of atoms and molecules in the system, producing a temperature increase. Joule’s contributions to thermodynamics were so significant that the SI unit of energy was named after him.

Increasing internal energy by heat transfer gives the same result as increasing it by doing work. Therefore, although a system has a well-defined internal energy, we cannot say that it has a certain “heat content” or “work content.” A well-defined quantity that depends only on the current state of the system, rather than on the history of that system, is known as a state variable . Temperature and internal energy are state variables. To sum up this paragraph, heat and work are not state variables .

Incidentally, increasing the internal energy of a system does not necessarily increase its temperature. As we’ll see in the next section, the temperature does not change when a substance changes from one phase to another. An example is the melting of ice, which can be accomplished by adding heat or by doing frictional work, as when an ice cube is rubbed against a rough surface.

Temperature Change and Heat Capacity

We have noted that heat transfer often causes temperature change. Experiments show that with no phase change and no work done on or by the system, the transferred heat is typically directly proportional to the change in temperature and to the mass of the system, to a good approximation. (Below we show how to handle situations where the approximation is not valid.) The constant of proportionality depends on the substance and its phase, which may be gas, liquid, or solid. We omit discussion of the fourth phase, plasma, because although it is the most common phase in the universe, it is rare and short-lived on Earth.

We can understand the experimental facts by noting that the transferred heat is the change in the internal energy, which is the total energy of the molecules. Under typical conditions, the total kinetic energy of the molecules K total K total is a constant fraction of the internal energy (for reasons and with exceptions that we’ll see in the next chapter). The average kinetic energy of a molecule K ave K ave is proportional to the absolute temperature. Therefore, the change in internal energy of a system is typically proportional to the change in temperature and to the number of molecules, N . Mathematically, Δ U ∝ Δ K total = N K ave ∝ N Δ T Δ U ∝ Δ K total = N K ave ∝ N Δ T The dependence on the substance results in large part from the different masses of atoms and molecules. We are considering its heat capacity in terms of its mass, but as we will see in the next chapter, in some cases, heat capacities per molecule are similar for different substances. The dependence on substance and phase also results from differences in the potential energy associated with interactions between atoms and molecules.

Heat Transfer and Temperature Change

A practical approximation for the relationship between heat transfer and temperature change is:

where Q is the symbol for heat transfer (“quantity of heat”), m is the mass of the substance, and Δ T Δ T is the change in temperature. The symbol c stands for the specific heat (also called “ specific heat capacity ”) and depends on the material and phase. In the SI system, the specific heat is numerically equal to the amount of heat necessary to change the temperature of 1.00 1.00 kg of mass by 1.00 ° C 1.00 ° C . The SI unit for specific heat is J/ ( kg × K ) J/ ( kg × K ) or J/ ( kg × °C ) J/ ( kg × °C ) . (Recall that the temperature change Δ T Δ T is the same in units of kelvin and degrees Celsius.)

Values of specific heat must generally be measured, because there is no simple way to calculate them precisely. Table 1.3 lists representative values of specific heat for various substances. We see from this table that the specific heat of water is five times that of glass and 10 times that of iron, which means that it takes five times as much heat to raise the temperature of water a given amount as for glass, and 10 times as much as for iron. In fact, water has one of the largest specific heats of any material, which is important for sustaining life on Earth.

The specific heats of gases depend on what is maintained constant during the heating—typically either the volume or the pressure. In the table, the first specific heat value for each gas is measured at constant volume, and the second (in parentheses) is measured at constant pressure. We will return to this topic in the chapter on the kinetic theory of gases.

In general, specific heat also depends on temperature. Thus, a precise definition of c for a substance must be given in terms of an infinitesimal change in temperature. To do this, we note that c = 1 m Δ Q Δ T c = 1 m Δ Q Δ T and replace Δ Δ with d :

Except for gases, the temperature and volume dependence of the specific heat of most substances is weak at normal temperatures. Therefore, we will generally take specific heats to be constant at the values given in the table.

Example 1.5

Calculating the required heat.

• Calculate the temperature difference: Δ T = T f − T i = 60.0 ° C . Δ T = T f − T i = 60.0 ° C .
• Calculate the mass of water. Because the density of water is 1000 kg/m 3 1000 kg/m 3 , 1 L of water has a mass of 1 kg, and the mass of 0.250 L of water is m w = 0.250 kg m w = 0.250 kg .
• Calculate the heat transferred to the water. Use the specific heat of water in Table 1.3 : Q w = m w c w Δ T = ( 0.250 kg ) ( 4186 J/kg ° C ) ( 60.0 ° C ) = 62.8 kJ . Q w = m w c w Δ T = ( 0.250 kg ) ( 4186 J/kg ° C ) ( 60.0 ° C ) = 62.8 kJ .
• Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in Table 1.3 : Q Al = m A1 c A1 Δ T = ( 0.500 kg ) ( 900 J/kg ° C ) ( 60.0 ° C ) = 27.0 kJ . Q Al = m A1 c A1 Δ T = ( 0.500 kg ) ( 900 J/kg ° C ) ( 60.0 ° C ) = 27.0 kJ .
• Find the total transferred heat: Q Total = Q W + Q Al = 89.8 kJ . Q Total = Q W + Q Al = 89.8 kJ .

Significance

Example 1.6 illustrates a temperature rise caused by doing work. (The result is the same as if the same amount of energy had been added with a blowtorch instead of mechanically.)

Example 1.6

Calculating the temperature increase from the work done on a substance.

Calculate the temperature increase of 10 kg of brake material with an average specific heat of 800 J/kg · °C 800 J/kg · °C if the material retains 10% of the energy from a 10,000-kg truck descending 75.0 m (in vertical displacement) at a constant speed.

Because the kinetic energy of the truck does not change, conservation of energy tells us the lost potential energy is dissipated, and we assume that 10% of it is transferred to internal energy of the brakes, so take Q = M g h / 10 Q = M g h / 10 . Then we calculate the temperature change from the heat transferred, using

where m is the mass of the brake material. Insert the given values to find

In a common kind of problem, objects at different temperatures are placed in contact with each other but isolated from everything else, and they are allowed to come into equilibrium. A container that prevents heat transfer in or out is called a calorimeter , and the use of a calorimeter to make measurements (typically of heat or specific heat capacity) is called calorimetry .

We will use the term “calorimetry problem” to refer to any problem in which the objects concerned are thermally isolated from their surroundings. An important idea in solving calorimetry problems is that during a heat transfer between objects isolated from their surroundings, the heat gained by the colder object must equal the heat lost by the hotter object, due to conservation of energy:

We express this idea by writing that the sum of the heats equals zero because the heat gained is usually considered positive; the heat lost, negative.

Example 1.7

Calculating the final temperature in calorimetry.

• Use the equation for heat transfer Q = m c Δ T Q = m c Δ T to express the heat transferred from the pan in terms of the mass of the pan, the specific heat of aluminum, the initial temperature of the pan, and the final temperature: Q hot = m A1 c A1 ( T f − 150 ° C ) . Q hot = m A1 c A1 ( T f − 150 ° C ) .
• Express the heat gained by the water in terms of the mass of the water, the specific heat of water, the initial temperature of the water, and the final temperature: Q cold = m w c w ( T f − 20.0 ° C ) . Q cold = m w c w ( T f − 20.0 ° C ) .
• Note that Q hot < 0 Q hot < 0 and Q cold > 0 Q cold > 0 and that as stated above, they must sum to zero: Q cold + Q hot = 0 Q cold = − Q hot m w c w ( T f − 20.0 ° C ) = − m A1 c A1 ( T f − 150 ° C ) . Q cold + Q hot = 0 Q cold = − Q hot m w c w ( T f − 20.0 ° C ) = − m A1 c A1 ( T f − 150 ° C ) .
• Bring all terms involving T f T f on the left hand side and all other terms on the right hand side. Solving for T f , T f , T f = m A1 c A1 ( 150 ° C ) + m w c w ( 20.0 ° C ) m A1 c A1 + m w c w , T f = m A1 c A1 ( 150 ° C ) + m w c w ( 20.0 ° C ) m A1 c A1 + m w c w , and insert the numerical values: T f = ( 0.500 kg ) ( 900 J/kg ° C ) ( 150 ° C ) + ( 0.250 kg ) ( 4186 J/kg ° C ) ( 20.0 ° C ) ( 0.500 kg ) ( 900 J/kg ° C ) + ( 0.250 kg ) ( 4186 J/kg ° C ) = 59.1 ° C . T f = ( 0.500 kg ) ( 900 J/kg ° C ) ( 150 ° C ) + ( 0.250 kg ) ( 4186 J/kg ° C ) ( 20.0 ° C ) ( 0.500 kg ) ( 900 J/kg ° C ) + ( 0.250 kg ) ( 4186 J/kg ° C ) = 59.1 ° C .

Check Your Understanding 1.3

If 25 kJ is necessary to raise the temperature of a rock from 25 °C to 30 ° C, 25 °C to 30 ° C, how much heat is necessary to heat the rock from 45 °C to 50 ° C 45 °C to 50 ° C ?

Example 1.8

Temperature-dependent heat capacity.

We solve this equation for Q by integrating both sides: Q = m ∫ T 1 T 2 c d T . Q = m ∫ T 1 T 2 c d T .

Then we substitute the given values in and evaluate the integral:

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Principle verification of the calibration light source subsystem for the calorimeter in herd experiment

• Original Paper
• Published: 09 September 2024

Cite this article

• Qianjun Chen 1 , 2 ,
• Xin Liu 2 ,
• Zhigang Wang 2 ,
• Zheyuan Zhang 2 ,
• Li Zhang 2 ,
• Yunpeng Lu   ORCID: orcid.org/0000-0001-9070-5458 2 ,
• Guozheng Nie 1 ,
• Kejun Zhu 2 &
• Yongwei Dong 2  

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The High energy cosmic-radiation detection facility (HERD) is a space astronomy and particle astrophysics experiment planned to be installed on the Chinese space station. The core detector of HERD, a three-dimensional imaging calorimeter, uses LYSO crystals as the detection medium, with wavelength-shifting fibers readout by image intensifier and CMOS imaging chips. In order to monitor the performance of the calorimeter in orbit, a highly integrated calibration light source subsystem is required to test the pulse linearity of the readout system, including the optical fibers.

To meet this critical requirement, a Calibration Light Source Subsystem was proposed based on the dual-light-source method, and a test circuit was set up to prove the principle.

Results and conclusions

Using a photomultiplier tube and waveform sampling module, the entire system’s linear response range was measured, which is equivalent to 1–900 Minimum Ionizing Particles (MIPs), with a linearity better than ± 5%. After adding an attenuator (transmittance of 6%) to the front end of the photomultiplier tube, the linear response range expands to 14–28,000 MIPs with a linearity better than ± 5%. This result demonstrated the significant contribution of the PMT’s saturation to the linearity measurement. The addition of the attenuator reduces the adverse effects of PMT saturation, and therefore the linearity of the Calibration Light Source Subsystem itself is better than ± 5% within the required dynamic range. In addition, testing of a single light source showed that long-term stability can be maintained within a range of ± 0.5% under conditions of 10 – 20 °C. The calibration method described in this paper has the characteristics of high integration and independent control of channels, providing an efficient and flexible calibration solution for the readout of calorimeter in HERD experiment.

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Acknowledgements

This work is supported by the National Natural Science Foundation of China (Grant No. 12027803).

National Natural Science Foundation of China,12027803,Yongwei DONG.

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Hunan University of Science and Technology, Xiangtan, 411201, China

Qianjun Chen & Guozheng Nie

Institute of High Energy Physics, Chinese Academy of Sciences(CAS), Beijing, 100049, China

Qianjun Chen, Xin Liu, Zhigang Wang, Zheyuan Zhang, Li Zhang, Yunpeng Lu, Kejun Zhu & Yongwei Dong

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Chen, Q., Liu, X., Wang, Z. et al. Principle verification of the calibration light source subsystem for the calorimeter in herd experiment. Radiat Detect Technol Methods (2024). https://doi.org/10.1007/s41605-024-00492-7

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Published : 09 September 2024

DOI : https://doi.org/10.1007/s41605-024-00492-7

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