Specific Heat Capacity | Chemisode Simulations

Practical setup

Aim: determine the specific heat capacity of vegetable oil by comparing its temperature change with water heated on the same hotplate.

m(water) / g

m(oil) / g

Room temp / °C

Thermometer mode

Example hypothesis: Vegetable oil will have a lower specific heat capacity than water. Therefore, when equal masses are heated under similar conditions, vegetable oil should have a larger temperature increase than water.

Logbook score

0 XP

Earn XP for accurate readings, graph plotting, extrapolation, calculations and MC questions. Hints reduce XP slightly.

Draft

Hypothesis

Logbook checklist

• Start the experiment to generate a new data set.

• Read both thermometers as the experiment runs.

• Fill in the table while the slider or auto-run changes the time.

• After the experiment ends, check your thermometer accuracy.

• Plot both graphs, extrapolate the cooling line, then complete the calculations.

Virtual laboratory and results table

Current time: 0 s Phase: Ready

Water

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Vegetable oil

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Speed

Time slider 0 s

Start the experiment from the setup page. Use the slider or smooth auto-run to move through the practical.

Zoomed water thermometer

Read the scale carefully between 20 and 90 °C.

Zoomed oil thermometer

The zoomed view makes reading easier on mobile and desktop.

Result entry

The whole table is shown immediately so students can record all readings as the experiment runs. Accuracy is checked only after the experiment finishes.

Waiting for completion

Plot your own graph

Choose a substance, then tap/click the graph to place points. Use Delete to switch into eraser mode and remove individual points without clearing the rest.

Selected: water

Water Vegetable oil Revealed answer

No graph marked yet.

Correct graph shown. The dotted lines and open circles are the correct water and oil data points. Your plotted points remain visible so you can compare them.

Results table reference

This is a read-only copy of your recorded values, kept beside the graph for easier plotting.

Extrapolate from your plotted graph

This view reuses the same points and the same time scale as your Graph page. The y-axis stays at 0 s, where heating began. The dashed vertical line marks when the beakers were removed from the hotplate. Draw a best-fit line through each cooling section and extend it backwards to the y-axis. The temperature where the line crosses the y-axis is the extrapolated final temperature for that liquid.

Selected: water line

Your water points Your oil points Revealed best-fit lines

Corrected final water temperature / °C

Corrected final oil temperature / °C

Draw a best-fit line for both water and oil.

Read the temperatures where your two cooling best-fit lines meet the original y-axis at 0 s.

Solution: The green dashed best-fit lines run through the cooling data and are extended backwards across the original graph to the y-axis at 0 s. Their y-intercepts are the extrapolated final water and oil temperatures.

Why extrapolate?

Once a beaker is removed from the hotplate, it begins losing energy to its surroundings. Keep the original time scale, draw a best-fit line through the cooling data, and extend it backwards to the original y-axis at 0 s.

Stepped calculations

Use your initial and corrected final temperatures. The program checks each step separately.

Step 1: Calculate ΔT for water

ΔT(water) = T(final, corrected) − T(initial)

ΔT(water) / °C

Step 2: Calculate q for water

q = m × c × ΔT, using c(water) = 4.18 J g⁻¹ °C⁻¹

q(water) / J

Step 3: Calculate ΔT for oil

Use the initial oil temperature and the extrapolated oil temperature from the y-axis intercept.

ΔT(oil) / °C

Step 4: Calculate c for oil

Assume q(oil) = q(water), then c(oil) = q ÷ (m × ΔT)

c(oil) / J g⁻¹ °C⁻¹

No calculation checked yet.

Formula bank

• q = m × c × ΔT

• c(water) = 4.18 J g⁻¹ °C⁻¹

• c(oil) = q ÷ (m × ΔT)

Multiple choice analysis questions

Complete your logbook

Conclusion

Finish your work to receive a final summary.

Sample conclusion: Vegetable oil had a lower specific heat capacity than water because its temperature increased more when both liquids were heated under similar conditions. Extrapolating the cooling data improved the estimated final temperature and therefore improved the calculation of energy supplied. Experimental uncertainty remains because heat was lost to the surroundings and the two beakers may not have received exactly the same energy.

Teacher-style reflection prompts

• Which liquid had the larger temperature increase?

• How did extrapolation improve the result?

• What sources of error still remain?