Finding the specific heat capacity of a solid
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Finding the specific heat capacity of a solid

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this presentation explains about a simple practical which can be used to find the specific heat capacity of a given solid. The presentation also explains about some special definitions that have to be ...

this presentation explains about a simple practical which can be used to find the specific heat capacity of a given solid. The presentation also explains about some special definitions that have to be used when considering about the specific heat capacity of a substance.

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Finding the specific heat capacity of a solid Finding the specific heat capacity of a solid Presentation Transcript

  • FINDING THE SPECIFIC HEAT CAPACITY OF A SOLID See more at:  Facebook – https://www.facebook.com/AdityaAbeysinghePresentations  Slideshare - slideshare.net/adityaabeysinghe  Wordpress - adityaabeysinghepresentations.wordpress.com/abeysinghe -foundation/
  • SEE THE VIDEO FORMAT OF THIS PRESENTATION AT: https://www.youtube.com/w atch?v=FI_c6NDUomw See more of my videos at : https://www.youtube.com/channel/UCVF Ss7LUN4DSr0a4kkGt4Ag
  • The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius. So, if the mass of the object whose specific heat capacity is calculated is m, the specific heat capacity is c, then the heat needed to raise the temperature of the system by θ is Q = mcθ SPECIFIC HEAT CAPACITY OF A SOLID
  • EXPERIMENT stirer Thermometer External beaker Calorimeter Objects whose specific heat capacity is to be measured
  • Let the mass of calorimeter + stirer = m1 , mass of calorimeter + stirer + water = m2, initial temperature of the calorimeter = θ1 , Total mass of the objects = m3 , initial temperature of the objects = θ2 and Final temperature of the system = θ THEORY/PRINCIPLE
  • Furthermore let’s take the specific heat capacity of the calorimeter to be C, that of the water to be CW and that of the objects to be C0. Assuming that the heat loss to the environment as negligible, Heat emitted by the objects = Heat gained by (water + calorimeter) Therefore, m3C0(θ2 – θ ) = { m1C + (m2 – m1)CW}{θ – θ1} Thus, C0 = [{ m1C + (m2 – m1)CW}{θ – θ1}] / [m3 (θ2 – θ ) ] By this relationship we can find the specific heat capacity of the objects.
  • 1. Measure the weight of the calorimeter and the stirer. 2. Pour water to the calorimeter so that water fills about ⅓ of the calorimeter’s volume. 3. Measure the total weight of the objects to be about thrice the weight of the total volume of water 4. Heat the objects using a Nicholson’s heater and record their initial temperature. 5. Introduce the objects to the calorimeter + water system and stir the objects. 6. Record the final temperature of the system METHOD
  • 1. Transfer of heat to the environment during the experiment poses diffculties for the successful completion of the experiment. This can be minimized by, (i) To prevent heat loss from conduction, the calorimeter should be covered with heat insulators (ii) To prevent heat loss from convection and vaporization, the calorimeter should be closed with an insulating lid (iii) To prevent heat loss from radiation, the calorimeter surface should be polished. IMPORTANT POINTS
  • 2. For the quick transmission of heat from the objects to water, small pieces of the objects should be used instead of large pieces. 3. The action of a Nicholson heater is as follows: To calorimeter Case which can be lifted to free the objects Thermometer Water The objects Heat Water Introduce the objects to the heater. Then water is introduced to the heater through the upper inlet. Then heat the system. Observe the temperature change while heating. Do not let the temperature go past 100° C as when introduced to water, some water particles may evaporate. Then lift the stop case and free the objects towards the calorimeter.