The enthalpy change of combustion, $\Delta H_c$, of butane, $\text{C}_4\text{H}_{10}$, can be determined using the apparatus shown in Fig. 2.1. The following steps are carried out. step 1 Use a $500\text{ cm}^3$ measuring cylinder to transfer $320\text{ cm}^3$ of water into a metal can. step 2 Place a thermometer into the water. Record the initial temperature of the water in the metal can. step 3 Weigh the gas container with burner and record the initial mass. step 4 Set up…

(a) 57.3 and 2.10 · (b) 76600 J · (c) -2120 kJ mol-1 · (d) Stir the water · (e) 0.78% · (f) Use a lid · (g)(i) Percentage error increases · (g)(ii) Less heat loss

Chemistry 9701/52 — May/June 2025

Cambridge A-Level · Planning, Analysis and Evaluation · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Analysis, Conclusions and Evaluation · Planning

Q1MediumPlanningAnalysis, Conclusions and Evaluation

Activated carbon is a form of carbon with the ability to adsorb solutes from solutions. The adsorption of ethanedioic acid molecules, $(\text{COOH})_2$ , from dilute ethanedioic acid, $(\text{COOH})_2(\text{aq})$ , onto the surface of activated carbon is studied in a series of experiments.

Before beginning the experiment, $500.0\text{ cm}^3$ of $0.139\text{ mol dm}^{-3}$ $(\text{COOH})_2(\text{aq})$ , solution X, is made. The activated carbon is placed in an oven at $120\text{ }^\circ\text{C}$ for three hours.

The experimental procedure involves the following steps.

step 1 Prepare $100.0\text{ cm}^3$ of $0.00556\text{ mol dm}^{-3}$ $(\text{COOH})_2(\text{aq})$ from solution X.
step 2 Transfer $50.0\text{ cm}^3$ of the $0.00556\text{ mol dm}^{-3}$ $(\text{COOH})_2(\text{aq})$ into a conical flask.
step 3 Add $0.500\text{ g}$ of activated carbon to the conical flask and start a stopwatch.
step 4 Shake the flask and immediately remove a $5.0\text{ cm}^3$ sample from the flask. Determine the mass of non-adsorbed $(\text{COOH})_2$ in the sample using chromatography.
step 5 Repeat step 4 at suitable time intervals until there is no further change in the mass of non-adsorbed $(\text{COOH})_2$ in the sample.

(a)

Solid ethanedioic acid has the formula $(\text{COOH})_2\cdot 2\text{H}_2\text{O}$ .

(i)

Calculate the mass of $(\text{COOH})_2\cdot 2\text{H}_2\text{O}(\text{s})$ required to prepare $500.0\text{ cm}^3$ of $0.139\text{ mol dm}^{-3}$ $(\text{COOH})_2(\text{aq})$ , solution X.

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(ii)

The mass of $(\text{COOH})_2\cdot 2\text{H}_2\text{O}(\text{s})$ calculated in (a)(i) is placed into a $100\text{ cm}^3$ beaker.

Describe the steps taken to prepare $500.0\text{ cm}^3$ of $0.139\text{ mol dm}^{-3}$ $(\text{COOH})_2(\text{aq})$ .

Give the name and capacity of any apparatus used.

Write your answer using a series of numbered steps.

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(b)

$(\text{COOH})_2\cdot 2\text{H}_2\text{O}(\text{s})$ is corrosive. Other than wearing eye protection and a lab coat, state one safety precaution that should be taken when using $(\text{COOH})_2\cdot 2\text{H}_2\text{O}(\text{s})$ .

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(c)

Suggest why the activated carbon is placed in the oven before use.

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(d)

In step 1, solution X is diluted to a concentration of $0.00556\text{ mol dm}^{-3}$ .

Calculate the volume of solution X needed to make $100.0\text{ cm}^3$ of $0.00556\text{ mol dm}^{-3}$ $(\text{COOH})_2(\text{aq})$ .

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(e)

Identify the most appropriate piece of equipment to transfer $50.0\text{ cm}^3$ of diluted $(\text{COOH})_2(\text{aq})$ in step 2.

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(f)

Suggest a reason why the conical flask is shaken in step 4.

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(g)

The results of the experiment are shown in Table 1.1.

Table 1.1

time, $t/\text{min}$	mass of non-adsorbed $(\text{COOH})_2$ in $5.0\text{ cm}^3$ sample / mg	mass of $(\text{COOH})_2$ adsorbed by activated carbon in $5.0\text{ cm}^3$ sample / mg
0	2.50	0.00
5	1.90
15	1.52
30	1.43
45	0.99
60	0.88
90	0.81
120	0.80
150	0.80

(i)

Complete Table 1.1.

1M Similar

DifficultyEasy

Worked solution

Working

Mass of $(\text{COOH})_2$ adsorbed = initial mass ( $2.50\text{ mg}$ ) $-$ mass of non-adsorbed $(\text{COOH})_2$ .

At $t = 5$ : $2.50 - 1.90 = 0.60\text{ mg}$
At $t = 15$ : $2.50 - 1.52 = 0.98\text{ mg}$
At $t = 30$ : $2.50 - 1.43 = 1.07\text{ mg}$
At $t = 45$ : $2.50 - 0.99 = 1.51\text{ mg}$
At $t = 60$ : $2.50 - 0.88 = 1.62\text{ mg}$
At $t = 90$ : $2.50 - 0.81 = 1.69\text{ mg}$
At $t = 120$ : $2.50 - 0.80 = 1.70\text{ mg}$
At $t = 150$ : $2.50 - 0.80 = 1.70\text{ mg}$

Answer

time, $t/\text{min}$	mass of non-adsorbed $(\text{COOH})_2$ in $5.0\text{ cm}^3$ sample / mg	mass of $(\text{COOH})_2$ adsorbed by activated carbon in $5.0\text{ cm}^3$ sample / mg
0	2.50	0.00
5	1.90	0.60
15	1.52	0.98
30	1.43	1.07
45	0.99	1.51
60	0.88	1.62
90	0.81	1.69
120	0.80	1.70
150	0.80	1.70

Final answer

0.60, 0.98, 1.07, 1.51, 1.62, 1.69, 1.70, 1.70

Detailed explanation

Background Concept

By conservation of mass, the total amount of ethanedioic acid initially present in the $5.0\text{ cm}^3$ sample is $2.50\text{ mg}$ (the value at $t = 0$ ). Any reduction in the mass remaining in solution corresponds to the mass that has been adsorbed onto the activated carbon:

\text{mass adsorbed} = m_{\text{initial}} - m_{\text{non-adsorbed}} = 2.50 - m_{\text{non-adsorbed}}

Techniques used

calculate derived values by subtractioncomplete data table with correct precision

(ii)

Plot a graph on the grid in Fig. 1.1 to show the relationship between mass of $(\text{COOH})_2$ adsorbed by activated carbon in $5.0\text{ cm}^3$ sample and time, $t$ . Use a cross ( $\times$ ) to plot each data point.

Draw a curved line of best fit.

Fig. 1.1

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(iii)

Identify the independent variable.

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(iv)

Suggest one variable that needs to be controlled that is not stated in the experimental procedure.

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(v)

Circle the one point on the graph that you consider to be most anomalous.

Explain the error in timing that may have led to this anomalous point.

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(h)

The experiment is repeated using different concentrations of $(\text{COOH})_2(\text{aq})$ prepared in step 1. The results are shown in Table 1.2.

Table 1.2

1	2	3	4	5
$m_i/\text{mg}$	$m_e/\text{mg}$	$(m_i - m_e)/\text{mg}$	$q_e/\text{mg g}^{-1}$	$\frac{m_e}{q_e}/\text{g}$
2.5	0.80	1.70	34.0	0.0235
5.0	2.10	2.90	58.0	0.0362
7.5	4.03	3.47	69.4	0.0581
10.0	6.41	3.59	71.8	0.0893
12.5	8.87	3.63	72.6	0.1222

$m_i$ is the initial mass of $(\text{COOH})_2$ in $5.0\text{ cm}^3$ sample.
$m_e$ is the mass of non-adsorbed $(\text{COOH})_2$ in the final $5.0\text{ cm}^3$ sample at the end of step 5.
$q_e$ is the mass of $(\text{COOH})_2$ adsorbed per gram of activated carbon.

(i)

A plot of $\frac{m_e}{q_e}$ against $m_e$ gives the line of best fit shown in Fig. 1.2.

Use Fig. 1.2 to determine the gradient of the line of best fit.

State the coordinates of both points you used in your calculation.

Include units in your answer.

Fig. 1.2

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(ii)

The adsorption parameter, $q_0$ , is the maximum mass of $(\text{COOH})_2$ adsorbed per gram of activated carbon.

The equation of the line of best fit plotted in Fig. 1.2 is shown.

\frac{m_e}{q_e} = \frac{m_e}{q_0} + \text{constant}

Use the gradient determined in (h)(i) to calculate the adsorption parameter, $q_0$ .

[If you were unable to determine an answer to (h)(i), then use the value 0.0136 for the gradient. This is not the correct answer.]

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(iii)

The experiment is repeated and the value of $q_0$ is calculated to be $78.1\text{ mg g}^{-1}$ . The total percentage error from the experimental procedure is $6.5\%$ .

The data book value of $q_0$ is $86.0\text{ mg g}^{-1}$ .

Use this information to determine whether the error in the repeated experiment could be accounted for by experimental errors or is caused by other factors.

Show your working.

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Q2Medium-EasyAnalysis, Conclusions and Evaluation

The enthalpy change of combustion, $\Delta H_c$ , of butane, $\text{C}_4\text{H}_{10}$ , can be determined using the apparatus shown in Fig. 2.1.

Fig. 2.1

The following steps are carried out.

step 1 Use a $500\text{ cm}^3$ measuring cylinder to transfer $320\text{ cm}^3$ of water into a metal can.
step 2 Place a thermometer into the water. Record the initial temperature of the water in the metal can.
step 3 Weigh the gas container with burner and record the initial mass.
step 4 Set up the apparatus as shown in Fig. 2.1.
step 5 Light the burner and allow the flame to heat the water in the metal can for three minutes.
step 6 Switch off the burner and record the maximum temperature reached.
step 7 When cool, reweigh the gas container with burner and record the final mass.

The results are shown in Table 2.1.

Table 2.1

initial temperature of water / $^\circ\text{C}$	maximum temperature of water / $^\circ\text{C}$	change in temperature of water, $\Delta T / ^\circ\text{C}$	initial mass of gas container with burner / g	final mass of gas container with burner / g	mass of butane burned / g
19.3	76.6		183.56	181.46

(a)

Complete Table 2.1. Record your answers to the appropriate number of decimal places.

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DifficultyEasy

Worked solution

Answer

Change in temperature of water, $\Delta T = 76.6 - 19.3 = 57.3 \, ^\circ\text{C}$
Mass of butane burned $= 183.56 - 181.46 = 2.10 \, \text{g}$

Final answer

57.3 and 2.10

Detailed explanation

Background Concept

When recording data from instruments, the precision of a calculated result cannot exceed the precision of the raw measurements. For addition and subtraction, the rule is to round the final answer to the same number of decimal places as the measurement with the fewest decimal places. This preserves the uncertainty inherent in the original readings.

Understanding the Question

The question asks to complete Table 2.1 by calculating the temperature change ( $\Delta T$ ) and the mass of butane burned. The raw data provided are initial and final temperatures (to 1 decimal place) and initial and final masses (to 2 decimal places).

Approach

Subtract the initial reading from the final reading for both temperature and mass. Ensure the final answers are recorded to the correct number of decimal places as dictated by the source data.

Step-by-Step Reasoning

Temperature change ( $\Delta T$ ):
The maximum temperature is $76.6 \, ^\circ\text{C}$ and the initial temperature is $19.3 \, ^\circ\text{C}$ .
$\Delta T = 76.6 - 19.3 = 57.3 \, ^\circ\text{C}$
Both readings are to 1 decimal place, so the answer is given to 1 decimal place.
Mass of butane burned:
The initial mass of the container with burner is $183.56 \, \text{g}$ and the final mass is $181.46 \, \text{g}$ .
$\text{Mass burned} = 183.56 - 181.46 = 2.10 \, \text{g}$
Both readings are to 2 decimal places, so the answer is given to 2 decimal places (the trailing zero is significant and must be kept).

Key Takeaways

Always match the decimal places of calculated differences to the least precise original measurement. Never drop trailing zeros after a decimal point, as they indicate precision (e.g., $2.1 \, \text{g}$ is less precise than $2.10 \, \text{g}$ ).

Common Mistakes

Recording $2.1 \, \text{g}$ instead of $2.10 \, \text{g}$ , losing a mark for incorrect significant figures/decimal places.
Calculating $181.46 - 183.56$ and getting a negative mass without taking the absolute value.

Things to Be Careful About

Ensure you subtract the smaller number from the larger number when finding a difference in mass or temperature to avoid negative values where a positive magnitude is expected. Keep the trailing zero in $2.10$ .

Techniques used

calculate temperature difference from readingscalculate mass difference from balance readings

(b)

Use the relationship $q = mc\Delta T$ to calculate the energy, $q$ , in J, gained by the water.

$1.00\text{ cm}^3$ of water has a mass of $1.00\text{ g}$ .

1M Similar

DifficultyEasy

Worked solution

Working

$q = m c \Delta T$

$q = 320 \times 4.18 \times 57.3 = 76603.2 \, \text{J}$

Answer

$76600 \, \text{J}$ (or $7.66 \times 10^4 \, \text{J}$ )

Final answer

76600 J

Detailed explanation

Background Concept

The heat energy ( $q$ ) gained or lost by a substance is calculated using the equation $q = mc\Delta T$ , where $m$ is the mass in grams, $c$ is the specific heat capacity in $\text{J g}^{-1} \text{K}^{-1}$ (or $\text{J g}^{-1} \, ^\circ\text{C}^{-1}$ ), and $\Delta T$ is the temperature change in Kelvin or Celsius. For water, the specific heat capacity $c$ is $4.18 \, \text{J g}^{-1} \text{K}^{-1}$ . The density of water is approximately $1.00 \, \text{g cm}^{-3}$ , meaning $1.00 \, \text{cm}^3$ of water has a mass of $1.00 \, \text{g}$ .

Understanding the Question

Calculate the energy gained by the water in joules. The volume of water is $320 \, \text{cm}^3$ , which converts to a mass of $320 \, \text{g}$ . The specific heat capacity and temperature change are known or calculated in previous parts.

Approach

Substitute the mass of water ( $320 \, \text{g}$ ), the specific heat capacity of water ( $4.18 \, \text{J g}^{-1} \text{K}^{-1}$ ), and the temperature change ( $57.3 \, \text{K}$ ) into the equation $q = mc\Delta T$ .

Step-by-Step Reasoning

Identify the values:
- $m = 320 \, \text{g}$ (since $1.00 \, \text{cm}^3 = 1.00 \, \text{g}$ )
- $c = 4.18 \, \text{J g}^{-1} \text{K}^{-1}$
- $\Delta T = 57.3 \, \text{K}$ (or $^\circ\text{C}$ , the magnitude is the same)
Calculate $q$ :
$q = 320 \times 4.18 \times 57.3 = 76603.2 \, \text{J}$
Round to appropriate significant figures. The given values ( $320$ , $4.18$ , $57.3$ ) generally have 3 significant figures, so the answer is rounded to 3 significant figures:
$q = 76600 \, \text{J} \quad \text{or} \quad 7.66 \times 10^4 \, \text{J}$

Key Takeaways

Always convert volume of water to mass using the given density ( $1 \, \text{g cm}^{-3}$ ). Remember that a temperature change in $^\circ\text{C}$ is numerically identical to a change in K.

Common Mistakes

Forgetting to convert $\text{cm}^3$ of water to grams (using $320$ instead of $320$ is fine here, but if density was different, it would matter).
Using the wrong specific heat capacity value. $4.18 \, \text{J g}^{-1} \text{K}^{-1}$ is the standard value expected in A-Level Chemistry unless otherwise given.
Forgetting to convert the final answer to kJ if the question asked for it (though this part asks for J).

Things to Be Careful About

Significant figures: $76603.2$ must be rounded to $76600$ (3 s.f.). Writing $76603$ or $76603.2$ may lose the accuracy mark depending on strictness, though mark schemes often allow unrounded intermediate values. Always carry unrounded values to the next part.

Techniques used

apply q = mc\Delta T formulaconvert volume of water to mass using density

(c)

Calculate the enthalpy change of combustion, $\Delta H_c$ , of butane, in $\text{kJ mol}^{-1}$ .

Give your answer to three significant figures.

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DifficultyMedium-Easy

Worked solution

Working

Molar mass of butane, $\text{C}_4\text{H}_{10} = (4 \times 12.0) + (10 \times 1.0) = 58.0 \, \text{g mol}^{-1}$

$\text{Moles of butane} = \frac{2.10}{58.0} = 0.0362 \, \text{mol}$

$\Delta H_c = \frac{-q}{n} = \frac{-76603.2}{0.0362} = -2116000 \, \text{J mol}^{-1}$

$\Delta H_c = -2116 \, \text{kJ mol}^{-1}$

Answer

$-2120 \, \text{kJ mol}^{-1}$

Final answer

-2120 kJ mol-1

Detailed explanation

Background Concept

The enthalpy change of combustion ( $\Delta H_c$ ) is the energy released when one mole of a substance burns completely in oxygen. It is calculated using $\Delta H = -q / n$ , where $q$ is the heat energy absorbed by the surroundings (water) in joules, and $n$ is the number of moles of fuel burned. The negative sign indicates that combustion is an exothermic process (energy is released from the system to the surroundings).

Understanding the Question

Calculate $\Delta H_c$ of butane in $\text{kJ mol}^{-1}$ to three significant figures. We have the energy gained by the water ( $q = 76603.2 \, \text{J}$ ) and the mass of butane burned ( $2.10 \, \text{g}$ ). We need to find the moles of butane and then the energy per mole.

Approach

Calculate the molar mass of butane ( $\text{C}_4\text{H}_{10}$ ).
Calculate the moles of butane burned using $n = m / M_r$ .
Calculate $\Delta H_c$ using $\Delta H_c = -q / n$ .
Convert from J to kJ and round to 3 significant figures.

Step-by-Step Reasoning

Molar mass of butane ( $\text{C}_4\text{H}_{10}$ ):
Using standard A-Level atomic masses ( $C = 12.0$ , $H = 1.0$ ):
$M_r = (4 \times 12.0) + (10 \times 1.0) = 48.0 + 10.0 = 58.0 \, \text{g mol}^{-1}$
Moles of butane burned:
$n = \frac{2.10}{58.0} = 0.036206... \, \text{mol} \approx 0.0362 \, \text{mol}$
Enthalpy change of combustion:
The energy released by the combustion is absorbed by the water, so $q_{\text{water}} = +76603.2 \, \text{J}$ . The energy change for the reaction is $-76603.2 \, \text{J}$ .
$\Delta H_c = \frac{-76603.2 \, \text{J}}{0.036206 \, \text{mol}} = -2115700 \, \text{J mol}^{-1}$
Convert to kJ mol $^{-1}$ and round:
$\Delta H_c = -2115.7 \, \text{kJ mol}^{-1}$
Rounding to 3 significant figures gives $-2120 \, \text{kJ mol}^{-1}$ .

Key Takeaways

Always include the negative sign for exothermic reactions. Convert J to kJ by dividing by 1000. Use unrounded intermediate values (like $0.036206...$ ) in the final calculation to avoid rounding errors.

Common Mistakes

Forgetting the negative sign (enthalpy change of combustion is always negative for exothermic reactions).
Forgetting to convert J to kJ (giving $-2120000$ instead of $-2120$ ).
Using the wrong molar mass (e.g., forgetting to multiply H by 10, giving $M_r = 50$ ).
Rounding intermediate values too early (e.g., using $n = 0.036$ gives $-2128$ , which is wrong).

Things to Be Careful About

The question specifically asks for three significant figures. $-2115.7$ rounded to 3 s.f. is $-2120$ . Ensure the trailing zero is not counted as significant incorrectly; scientific notation ( $-2.12 \times 10^3$ ) can avoid ambiguity, but $-2120$ is standard here. Remember that $\Delta H_c$ is defined per mole of fuel burned.

Techniques used

calculate moles from mass and molar masscalculate enthalpy change from energy and moles

(d)

Without changing the apparatus, suggest what should be done in step 6 before recording the maximum temperature reached to improve the experimental procedure.

1M Similar

DifficultyEasy

Worked solution

Answer

Stir the water before taking the temperature reading.

Final answer

Stir the water

Detailed explanation

Background Concept

In calorimetry experiments, heat is often applied locally (e.g., at the bottom of the can by the burner). Without stirring, the water near the heat source will be hotter than the water at the top or sides. This creates a temperature gradient, meaning a thermometer placed in the water may not record the true average maximum temperature.

Understanding the Question

The question asks for a procedural improvement in step 6 (recording the maximum temperature) without changing the apparatus. We need to ensure the temperature reading is accurate and represents the whole volume of water.

Approach

Stirring ensures that the heat is distributed evenly throughout the water, allowing the thermometer to record the true maximum average temperature.

Step-by-Step Reasoning

Action: Stir the water gently but continuously during heating, and especially just before recording the maximum temperature.
Reason: This eliminates temperature gradients and ensures the thermometer measures the average temperature of the entire water volume. Without stirring, the recorded maximum temperature might be lower than the actual average temperature, leading to an underestimation of $q$ and $\Delta H_c$ .

Key Takeaways

Stirring is a standard improvement in simple calorimetry to ensure thermal equilibrium throughout the liquid.

Common Mistakes

Suggesting to 'use a better thermometer' (this changes the apparatus).
Suggesting 'wait longer' (this doesn't help if the water is stratified).

Things to Be Careful About

The question specifies 'without changing the apparatus'. Therefore, suggestions like 'add a lid', 'use a copper can instead of steel', or 'use a digital thermometer' are incorrect. The answer must be an action the student can perform with the existing equipment.

Techniques used

identify procedural improvement for calorimetry

(e)

The $500\text{ cm}^3$ measuring cylinder has graduations every $5\text{ cm}^3$ .

Calculate the percentage error in the measurement of the volume of water.

Show your working.

1M Similar

DifficultyEasy

Worked solution

Working

Smallest division = $5 \, \text{cm}^3$
Maximum absolute error = $\frac{5}{2} = 2.5 \, \text{cm}^3$

$\text{Percentage error} = \frac{\text{absolute error}}{\text{measured value}} \times 100\%$

$\text{Percentage error} = \frac{2.5}{320} \times 100\% = 0.78125\%$

Answer

$0.78\%$

Final answer

0.78%

Detailed explanation

Background Concept

Every measuring instrument has a limit to its precision, defined by its smallest division. The maximum absolute error of a measurement is typically taken as half of the smallest division (assuming the user can estimate to the nearest half-division). The percentage error is a relative measure of this uncertainty, calculated as $(\text{absolute error} / \text{measured value}) \times 100\%$ .

Understanding the Question

Calculate the percentage error in measuring $320 \, \text{cm}^3$ of water using a $500 \, \text{cm}^3$ measuring cylinder with graduations every $5 \, \text{cm}^3$ .

Approach

Determine the maximum absolute error: half of $5 \, \text{cm}^3$ is $2.5 \, \text{cm}^3$ .
Apply the percentage error formula: $(2.5 / 320) \times 100$ .

Step-by-Step Reasoning

Absolute error:
The measuring cylinder has markings every $5 \, \text{cm}^3$ . The maximum error in reading the meniscus is half of this interval:
$\text{Absolute error} = \frac{5}{2} = 2.5 \, \text{cm}^3$
Percentage error:
$\% \text{ error} = \frac{2.5}{320} \times 100\% = 0.78125\%$
Rounding to 2 significant figures (standard for percentage errors in this context, or matching the precision of the inputs): $0.78\%$ .

Key Takeaways

Percentage error = $(\text{half smallest division} / \text{measured value}) \times 100$ . Larger measured values reduce the percentage error, which is why using a more precise instrument or measuring a larger volume improves accuracy.

Common Mistakes

Using the full smallest division ( $5 \, \text{cm}^3$ ) as the absolute error instead of half ( $2.5 \, \text{cm}^3$ ).
Forgetting to multiply by 100 to get a percentage.
Calculating the error for the wrong volume (e.g., using $500$ instead of $320$ ).

Things to Be Careful About

Ensure you use the volume actually measured ( $320 \, \text{cm}^3$ ), not the total capacity of the cylinder ( $500 \, \text{cm}^3$ ).

Techniques used

calculate percentage error from instrument precision

(f)

A student suggests that the value calculated in (c) is different from the actual value of $\Delta H_c$ of butane because of heat lost during the experiment. Suggest one change to the apparatus that would reduce the heat lost.

1M Similar

DifficultyEasy

Worked solution

Answer

Use a lid (or cover) on the metal can.

Final answer

Use a lid

Detailed explanation

Background Concept

In simple combustion calorimetry, a significant source of error is heat loss to the surroundings. Heat is lost via conduction through the can, convection from the water surface, and radiation. Minimizing these losses improves the accuracy of the calculated $\Delta H_c$ .

Understanding the Question

Suggest one change to the apparatus (not just the procedure) to reduce heat lost. The current apparatus is an open metal can on a tripod.

Approach

Identify the major pathway for heat loss. The open top of the can allows hot water vapor to escape and heat to be lost by convection and evaporation. Covering it with a lid traps the heat.

Step-by-Step Reasoning

Observation: The metal can is open at the top.
Heat loss mechanism: Evaporation of water and convection currents carry heat away from the water surface into the air.
Improvement: Place a lid (e.g., a piece of cardboard with holes for the thermometer and burner, or a dedicated calorimeter lid) on the metal can.
Effect: This reduces heat loss by evaporation and convection, ensuring more of the combustion energy is transferred to the water.

Key Takeaways

Lids, insulation (e.g., polystyrene cups instead of metal cans), and wind shields are standard apparatus improvements for calorimetry.

Common Mistakes

Suggesting 'stir the water' (this is a procedural change, not an apparatus change, and was the answer to part d).
Suggesting 'use a digital thermometer' (this improves precision of reading, not necessarily reducing heat loss).
Suggesting 'move the burner closer' (this increases heat loss to the air and may cause incomplete combustion).

Things to Be Careful About

The question asks for a change to the apparatus. 'Stir the water' is an action/procedure. 'Use a lid' is an additional piece of apparatus. Be precise with your wording.

Techniques used

propose apparatus modification to reduce heat loss

(g)

The experiment was repeated but the burner was switched off after only two minutes.

(i)

Suggest why this might contribute to a reduction in the accuracy of $\Delta H_c$ of butane.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The absolute error in temperature change (or mass of butane burned) remains the same, but the measured value is smaller, so the percentage error increases.

Final answer

Percentage error increases

Detailed explanation

Background Concept

Percentage error is inversely proportional to the measured value: $\% \text{ error} = (\text{absolute error} / \text{measured value}) \times 100$ . If the absolute error (determined by instrument precision) stays constant but the measured value decreases, the percentage error increases. This reduces the accuracy of the final calculated result.

Understanding the Question

The burner is switched off after 2 minutes instead of 3. This means less heat is transferred to the water, resulting in a smaller temperature change ( $\Delta T$ ) and less butane burned (smaller mass change). We need to explain why this reduces accuracy.

Approach

Consider the effect of smaller measured values on the percentage error calculation. The precision of the thermometer and balance doesn't change, so the absolute error is constant.

Step-by-Step Reasoning

Effect on measurements:
Heating for a shorter time (2 min vs 3 min) results in a smaller temperature rise (e.g., $\Delta T = 38 \, ^\circ\text{C}$ instead of $57.3 \, ^\circ\text{C}$ ) and less fuel burned (e.g., $1.40 \, \text{g}$ instead of $2.10 \, \text{g}$ ).
Effect on percentage error:
The absolute error in $\Delta T$ (e.g., $\pm 0.5 \, ^\circ\text{C}$ ) and mass (e.g., $\pm 0.01 \, \text{g}$ ) remains the same because the instruments haven't changed.
However, because the measured values ( $\Delta T$ and mass burned) are smaller, the ratio $(\text{absolute error} / \text{measured value})$ is larger.
$\text{Percentage error} = \frac{\text{absolute error}}{\text{smaller measured value}} \times 100\% \rightarrow \text{larger percentage}$
Conclusion:
The increased percentage error in the fundamental measurements ( $\Delta T$ and mass) propagates through the calculation, reducing the accuracy of the final $\Delta H_c$ value.

Key Takeaways

Smaller measured values lead to larger percentage errors if the instrument precision is unchanged. This is why calorimetry experiments often aim for a significant temperature rise (e.g., $20-30 \, ^\circ\text{C}$ ) to minimize percentage errors.

Common Mistakes

Saying 'the error is larger' (vague; specify percentage error increases because absolute error is constant while value is smaller).
Suggesting the thermometer is less accurate (instruments have fixed precision).

Things to Be Careful About

Distinguish between accuracy (closeness to true value, affected by systematic errors and percentage errors) and precision (reproducibility, affected by random errors). Here, the percentage error increases, reducing the reliability/accuracy of the result.

Techniques used

evaluate effect of smaller readings on percentage error

(ii)

Suggest why this might contribute to an increase in the accuracy of $\Delta H_c$ of butane.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

There is less time for heat to be lost to the surroundings, so a greater proportion of the energy released is transferred to the water.

Final answer

Less heat loss

Detailed explanation

Background Concept

In any real calorimetry experiment, heat is continuously lost to the surroundings (air, tripod, can) at a rate proportional to the temperature difference between the system and the surroundings. The longer the experiment runs, the more total heat is lost. This is a systematic error that causes the calculated $\Delta H_c$ to be less exothermic (smaller magnitude) than the true value.

Understanding the Question

Despite the increased percentage error (from part i), switching off the burner earlier (2 min instead of 3 min) might actually increase the accuracy of $\Delta H_c$ . We need to explain why.

Approach

Consider the cumulative effect of heat loss over time. A shorter experiment duration means less total time for heat to escape to the surroundings.

Step-by-Step Reasoning

Heat loss mechanism:
Heat is lost from the water, can, and apparatus to the surrounding air via convection, conduction, and radiation. This loss is continuous and accumulates over time.
Effect of shorter time:
By switching off the burner after 2 minutes instead of 3, the total duration of the experiment is reduced. Therefore, there is less total time for heat to be lost to the surroundings.
Effect on accuracy:
Although the percentage error in measurements increases (as discussed in part i), the reduction in cumulative heat loss means that a higher percentage of the energy released by combustion is actually absorbed by the water. This systematic error (heat loss) is reduced, which can improve the overall accuracy of the calculated $\Delta H_c$ value, potentially outweighing the increased percentage error from the smaller readings.

Key Takeaways

There is a trade-off in calorimetry: heating for too long increases cumulative heat loss (reducing accuracy), while heating for too short a time increases percentage errors in measurements (also reducing accuracy). An optimal temperature rise (usually $20-30 \, ^\circ\text{C}$ ) balances these two effects.

Common Mistakes

Saying 'less heat is produced' (the rate of heat production is the same; the total time is shorter).
Confusing this with part (i) and giving a reason about percentage error.

Things to Be Careful About

The question asks why it might contribute to an increase in accuracy. Focus on the reduction of systematic error (heat loss) rather than random error (percentage error).

Techniques used

evaluate effect of shorter heating time on heat loss