5. (a) Define the term \"specific heat capacity\". 14] (b) In a calorimetry expe

Question

5. (a) Define the term "specific heat capacity". 14] (b) In a calorimetry experiment, a 300 g mass of an unknown material, at a[10] temperature of 220°C, is placed in a copper calorimeter of mass 500 g, containing 1 litre of water. The specific heat capacity of copper is 380 J kg K, of water is 4190 J kg K1, and of the unknown mass is 350 J kg K1. The density of water is 1000 kg m3, the initial temperature measured in the experiment was 21°C, and at the end of the experiment was 55°C. What was the mass of the unknown material? (c) [6] A2 kg block of ice, initially at a temperature of 0°C, is heated at a rate of 600 W for 20 minutes. The ice melts completely. What is the temperature of the water at the end of the 20 minutes period of heating? The latent heat of melting of ice is 0.34 x106 J kg, and the specific heat capacity of water is 4190 J kg K1

Explanation / Answer

1) Specific Heat Capacity :
   It is the amount of heat that is required to raise the temperature of 1 kilogram mass by 1 degree centigrade (or 1 kelvin)
     
3) Ice is heated at the rate = 600W
Time of heating = 20 minutes = 20*60 sec
   Power = Energy/time
   Energy = heat = power * time
   = 600* (20*60)
   = 720000 joule
   Therefore Heat given to the ice = 720000 joule
   As the ice melted completely, so the energy given is utilized as latent heat and specific heat.
  
   Completed process = (At 0 degree centigrade ice melts [phase transmission]) + (melted liquid gets heated up)
   720000 = latent heat of ice + specific heat of water. ......(1)
   Latent heat of melting of ice at 0 C = 0.34 *106 (J/kg) * 2 kg
   = 680000 joule

   After the phase transmission, the melted liquid gets heated up
   Specific heat = m* Cp*(Tf - Ti )
   where m : mass of the water = mass of ice ( from conservation of mass principle)
   Cp : specific heat capacity of water
Ti & Tf : Initial & Final Temperatures respectively.
   = (2kg ) * (4190 J/Kg.K) * (Tf - 0)
from (1), it follows

720000 = 680000 + 8380 (Tf)
     Tf = (40000/8380)
   = 4.773 0C
After 20 minutes, the temperature of water = 4.773 0C

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