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Calculating specific heat capacities

Calculating specific heat capacities. Calculating the specific heat capacity of aluminium. V. A. 3 600.00. Current = 1.5A Pd across heater = 6V. Aluminium block mass 0.5kg. Initial temperature = 20 C Final temperature = 85C. Total electrical energy in E= IV Δ t = 1.5A x 6V x 3600s

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Calculating specific heat capacities

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  1. Calculating specific heat capacities

  2. Calculating the specific heat capacity of aluminium V A

  3. 3 600.00 Current = 1.5A Pd across heater = 6V Aluminium block mass0.5kg Initial temperature = 20 C Final temperature = 85C Total electrical energy in E= IVΔt = 1.5A x 6V x 3600s = 32 400J

  4. Q= mcΔθ So c = Q/mΔθ c = 32 400/ (0.5kg x 65K) =997Jkg-1K-1 Total electrical energy in 32 400J is taken as equal to Qthe heat energy transferred to the block The actual specific heat capacity of Al is 900 Jkg-1K-1 The discrepancy is due to heat loss. We can reduce heat losses in the normal way by insulating the system and use glycerine around the thermometer to improve thermal contact A rather clever way to compensate is to “correct for cooling” using a cooling correction factor which can be determined experimentally (this method is beyond the scope of the course)

  5. V A Liquid under test stirrer Still air reduces heat loss by conduction and convection

  6. Table of shc

  7. An aluminium saucepan of mass 1.2kg contains 0.5 kg of pure water. It is heated and the temperature of the pan and contents rise together as show in the graph below. Given the specific heat capacities of aluminium and water find the rate at which energy is supplied to both the pan and water.

  8. Aluminium has a specific heat capacity of 900Jkg-1K-1 Water has a specific heat capacity of 4 200Jkg-1K-1

  9. This aluminium saucepan has a mass of 1.2 kg Its heat capacity is 1.2 x 900 = 1080 JK-1 The water in it has a mass of 0.5kg The heat capacity of this mass of water is 0.5 x 4200 = 2100JK-1 So the total heat capacity of the aluminium and the water is 1080 + 2100 = 3180JK-1

  10. 30 C 10 min Degrees K per second Find the rate of temperature rise at the beginning of the heating process i.e. before too much evaporation of the liquid takes place which brings other factors into play.

  11. For the pan and water togetherHeat capacity = 3180JK-1Temperature rise=0.05Ks-1 Rate of energy change = Heat capacity x rate of temperature rise P.S. If you are doing this you should get around 230 W as the gradient of the graph that I have drawn is slightly different from that given in the question

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