SITEMAP   Thermal energy 2.5 Measuring specific heat capacity of a solid indirectly

UK GCSE level age ~14-16 ~US grades 9-10 Scroll down, take time to study content or follow links

Thermal energy: 2.5 Ways of measuring the heat capacity of a solid using an indirect method involving a preheating or pre-cooling the material in a liquid of known specific heat capacity

Doc Brown's Physics exam study revision notes

2.5a Ways of measuring the heat capacity of a solid using an indirect method (A)

(Procedures (iv)(A) and (iv)(B)  relate to part 2.2 calculation Q5 in the set of specific heat capacity questions above)

This method relies on the transfer between a brass weight's thermal energy store and the thermal energy store of water.

Illustration of procedure A

Method (procedure A)

A brass weight is accurately weighed and placed in beaker of ice-cold water (tap water plus lumps of ice)

This is left for some time and occasionally stirred, until the brass weight is at the same temperature as the water.

250 g of water is weighed into a beaker and heated until its about 80oC (no need to boil, increases hazard).

After a gentle stir, the temperature of the warm water is taken (T2), immediately before transfer.

The temperature (T1) of the iced-water is also taken just before transferring the brass weight from the iced-water to the warm water using tongs or thin wire and loop.

After transferring the cold brass weight to the warm water, allow time for the thermal heat transfer into the brass weight to complete and after a gentle stir read the final temperature (T3)

Results and calculation (A)

Typical results:

Mass of brass weight 200 g (0.200 kg), mass of water 250 g (0.250 kg), SHC water = 4180 J/kgoC

Initial temperature of iced water T1 = 0.5oC

Initial temperature of the warm water (T2) = 79.5oC

Final temperature of water/brass in 2nd beaker (T3) = 74.0oC (cooler, as brass weight absorbs heat)

E = Thermal energy transfer to brass weight = Thermal energy transfer from water (all in J)

E = m x c x ∆T = 0.200 x SHCbrass x (T3 - T1) = 0.250 x 4180 x (T2 - T3) = 5747.5

E = 0.200 x SHCbrass x (74.0 - 0.5) = 0.250 x 4180 x (79.5 - 74.0) = 5747.5

0.20 x SHCbrass x 73.5 = 5747.5

14.7 x SHCbrass = 5747.5

SHCbrass = 5747.5 / 14.7 = 391 J/kgoC (3 sf)

Improvements and sources of error (A)

(i) Could not be quite sure if the brass weight has completely cooled down to ~0oC (T1).

(ii) The hot water in the beaker is losing heat all the time giving a greater cooling effect than just from the brass weight - might be better to use a lower start temperature in the 2nd beaker e.g. 40-50oC.

2.5b Alternative procedure, but a similar methodology (B)

Illustration of procedure B

Method (procedure B)

You could do a similar experiment to (A) by putting the brass weight into hot water for some to heat up to ~80-90oC.

You could heat the water directly in a beaker to ~80-90oC.

Allow time for the brass weight to heat up.

After a careful gentle stir, measure the initial temperature of the brass weight (T1, ~90oC) in the hot water.

Measure the initial temperature of the cold water in the 2nd beaker (T2, ~20oC)

Pick up the brass weight with tongs or a thin wire and loop and transfer it to a beaker of water at room temperature.

Allow time for the thermal energy store transfers to take place.

After a careful gentle stir, measure the final temperature of the brass weight/water (T3) in the 'warmed up' water.

You can then do a similar calculation as above e.g.

Results and calculation (B)

Typical results:

Mass of brass weight 200 g (0.200 kg), mass of water 250 g (0.250 kg), SHC water = 4180 J/kgoC

Initial temperature of heated water/brass weight (T1) = 85.0oC

Initial temperature of the cold water (T2) = 20.5oC

Final temperature of water/brass in 2nd beaker(T3) = 24.9oC

E = Thermal energy transfer from hot brass weight = Thermal energy transfer to water (all in J)

E = m x c x ∆T = 0.200 x SHCbrass x (T1 - T3) = 0.250 x 4180 x (T3 - T2) = ?

E = 0.200 x SHCbrass x (85.0 - 24.9) = 0.250 x 4180 x (24.9 - 20.5) = 4598

0.20 x SHCbrass x 60.1 = 4598

12.02 x SHCbrass = 4598

SHCbrass = 4598 / 12.02 = 383 J/kgoC (3 sf)

Improvements and sources of error (B)

(i) Could not be quite sure if the brass weight has completely warmed up to ~80-90oC (T1)

(ii) Not absolutely sure the thermal energy transfer from the hot brass weight to the water is complete - if left too long the 2nd beaker of water will start to cool give a greater temperature change than should be measured.

(iii) The beaker absorbs some of the extra thermal energy transferred to the water's thermal energy store.

Keywords, phrases and learning objectives for measuring the specific heat capacity of a solid by an indirect method

Understand how to measure the heat capacity of solid using indirect method involving pre-cooling or preheating material in a liquid of known specific heat capacity like water.

WHAT NEXT?

INDEX for physics notes on specific heat capacity

email doc brown - comments - query?

BIG website, using the [SEARCH BOX] below, maybe quicker than navigating the many sub-indexes

for UK KS3 science students aged ~12-14, ~US grades 6-8

ChemistryPhysics for UK GCSE level students aged ~14-16, ~US grades 9-10

for pre-university age ~16-18 ~US grades 11-12, K12 Honors

Use your mobile phone in 'landscape' mode?

SITEMAP Website content © Dr Phil Brown 2000+. All copyrights reserved on Doc Brown's physics revision notes, images, quizzes, worksheets etc. Copying of website material is NOT permitted. Exam revision summaries and references to GCSE science course specifications are unofficial.

Using SEARCH some initial results may be ad links you can ignore - look for docbrown

 @import url(https://www.google.co.uk/cse/api/branding.css); ENTER specific physics words or courses e.g. topic, module, exam board, formula, concept, equation, 'phrase', homework question! anything of physics interest!  This is a very comprehensive Google generated search of my website

TOP OF PAGE