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Specific Heat Capacity and Latent Heat Experiments In Physics

The following experiment measures the specific heat capacity of a block of metal.


Before you start the experiment, you will need the equipment for the experiment. You will need:

  • 12V battery pack.
  • 5 wires (two black and three red preferably).
  • Ammeter.
  • Voltmeter.
  • Metal block (copper in our case).
  • Immersion heater.
  • Insulation to surround the metal block.
  • Stop clock.
The first step to the experiment is to set up a series circuit connecting the battery pack to ammeter and inversion heater and then back to the battery pack. You need to make sure that the battery pack is accurate and will need to use a Voltmeter to get the voltage coming out at exactly 12 volts. You need to also check all cables to make sure they work. Measure the mass of the metal block and record this.
The immersion heater needs to be placed into the metal block with the help of a bit of oil. Once you have done this, connect the Voltmeter in parallel to the series circuit so it is connecting only to the immersion heater.
It is important to fully cover your chosen metal block in insulation so that it doesn’t lose heat to the surroundings. You can now place the thermometer into the other whole in the metal block. The set up for the experiment is complete.
You need to record the first temperature reading of the metal block before the experiment. Once you have done this, you can turn the battery back on and start the stop clock. When recording the volts and amps, take an average for the duration of the experiment. 
After 5 minutes (300 seconds), turn the battery pack off but leave the circuit as it is for another minute. This is because the immersion heater after the battery pack is turned off will still be providing heat to the metal block. After a minute, record the temperature and find the temperature change (measuring temperature in degrees Celsius or kelvin). 
You now have all the information needed to work out the metal’s specific heat capacity. You have the voltage (V) and current (I) of the circuit and time (300s). Therefore, we can work out the power provided being E = VIT. Now we know E, the mass of the metal block and temperature change, we can work out the metal’s specific heat capacity through using the equation C = E / (m x △T). The answer will be in J/kg/K.

For our experiment, we gained the following information:
  • Mass of our copper block = 1.011kg
  • Voltage = 11.6V
  • Current = 4.9A
  • Power = VIT = 16, 824.6 J
  • Temperature change = 41K. 
The specific heat capacity (C) = E / (m x △T) = 16,824.6 / (1.011 x 41). This equals to 405.9 J/kg/K which is not far off the upper limit true value of 380-400 J/kg/K.

Specific Heat Capacity of Water

For the experiment to explore the specific heat capacity of water, we need:
  • Distilled water (250ml).
  • Aluminium calorimeter, lid and stirrer.
  • Thermometer.
  • 3-4V immersion heater.
  • 2 multimeters.
  • Electronic mass balance, 0.1g.
  • Stop clock.
  • Power supply.


  1. Set up the apparatus as shown in the diagram above but without the water.
  2. Check the heating element draws a current when 3-4V d.c. approximately is across the terminals. Use multimeters to measure this (Ammeter set to 10A d.c.)
  3. Measure the mass of the empty inner calorimeter and record this.
  4. Add enough water to bring the level to 1cm below the bayonet locking points on the lid.
  5. Measure the mass of the water and calorimeter and water and record this.
  6. Check that the bare wire element does not touch the thermometer when it is inserted.
  7. Carefully place the filled inner calorimeter inside the outer container so that it is supported by the ‘rim’.
  8. Measure the initial temperature of the water and record this.
  9. Ensure the 3-4V d.c. supply is connected and the ammeter is connected in series and the voltmeter in parallel (like the first experiment in this article).
  10. Prepare the stop clock.
  11. Begin the experiment by switching the power supply on and start the stop clock.
  12. Run the experiment until the temperature has risen by roughly 20 degrees Celsius (or kelvin: does not matter).
  13. Record the voltage and current readings every minute. 
  14. Switch the power supply off and record the time.
  15. Continue to watch the temperature rise until it peaks and then record this temperature.
  16. Leave the equipment to coll and consider your data – how will you use it to find a reliable and accurate answer for the specific heat capacity of water? Aluminium (which is what the calorimeter is made from) has a specific heat capacity of 880-937 J/kg/k at a temperature of 273-373K (0-100 degrees Celsius).
From doing the above experiment, we obtained the following data:
  • Mass of aluminium with water = 0.1477 kg.
  • Mass of aluminium without water = 0.0323 kg.
  • Therefore, mass of water = 0.1154 kg.
  • Starting temperature = 23 degrees Celsius.
  • Time taken to cause an rise in temperature of 8 degrees Celsius = 381s.
  • Voltage = 3.23V.
  • Current = 4A.
Below are the step by step equations that will result in us calculating the SHC of water:
  • If C = E / (m x △T), E = mC△T.
  • E = (mwater x Cwater x △T) + (maluminum x Caluminium x △T)
  • E / △T = (mwater x Cwater) + (maluminum x Caluminium)
  • (E/△T) – (maluminum x Caluminium) = (mwater x Cwater)
  • Cwater = ((E/△T) – (maluminum x Caluminium)) / mwater 
  • But, E = VIT.
  • Cwater = ((VIT/△T) – (maluminum x Caluminium)) / mwater 
This time, temperature has to be in kelvin and by using the SHC of aluminium and the temperature we were working at as 900, we produced a specific heat capacity, for water, of 5080 J/kg/K. The real answer is roughly 4200 so we are a bit out.

The electrical energy supplied will heat both the inner aluminium calorimeter and the water. This is why E = (mwater x Cwater x △T) + (maluminum x Caluminium x △T).

Specific Latent Heat

  • The specific latent heat is the energy that is absorbed by a chemical substance during a change of state. In this case, we are looking at how water obsorbs the energy from a kettle to enable it to vaporize.
  • T start = 21 degrees Celsius (294K). T finish = 100 degrees Celsius (373K).
  • Time to rolling boil = 155s.
  • Power = 2200W (kettle).
  • M (water and kettle) = 1.409kg. Mass of kettle = 0.596kg. At boiling temperature of water, mass = 1.385kg. After a rolling boil of 68 seconds, mass decreases by 0.050kg to 1.335kg.
  • Power = energy / time. therefore, energy = power x time.
  • The specific latent heat (L) = E / m. This produces the energy per unit mass.
  • We know that C (SHC) of water = 4200 J/kg/K.
  • L = (Pxt) / m = (2200×60) / 0.050 = 2.992×10^6 J/kg.
  • △E = mC△T = (1.409-0.596) x 4200 x 68 = 232.2×10^3J.
  • E = P x t = 2200 x 155 = 341,000J.
  • 230,000 / 341,000 = 67%. Only 67% of the energy produced from the kettle goes into the water to vaporize it.
  • 2.992×10^6 x 0.67 = 2005 KJ/kg.
  • For 1kg, E for vapourisation – 2.260×10^6J.
  • Molar mass of water = 0.018kg.
  • 0.018 x 2.260×10^6 = 40,680J/mol.
  • For one particle, 40,680/Na (6.02×10^23) = 6.76×10^-20J.
  • At 373K, KT is the average energy of the water molecules. 1.38×10^-23 x 373 = 5.15×10^-21 J. This roughly 12 times smaller than 6.76×10^-20J.


  • E = VIT.
  • To work out the specific heat capacity, C = E / m △T.
  • The specific latent heat (L) = E / m.

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