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Physics "O-level"

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Part 2 based on calculation of special Heat capacity and heat capacity.(must read part 1 to understand the symbols and signification)

Heat Capacity

How much heat energy is needed to heat 10kg of water?

Final temperature:25 Celsius
Initial temperature:24 Celsius
Note:the temperature has been raised only by 1 Celsius.

Heat Capacity=10x4200J
=42,000J/Celsius
=42kJ/Celsius


How much heat energy is needed to heat 5kg of mercury?

Final temperature:25 Celsius
Initial temperature:24 Celsius
Note:the temperature has been raised only by 1 Celsius.

Heat Capacity=5x140
=700J/Celsius



Calculate the quantity of heat needed to increase the temperature of 5 kg of iron from 225 Celsius to 345 Celsius(c of iron=400J/kg/k)
Ans;
Using formula
Q=mc fita
m=5kg
c=400J/kg/k
fita=345-225 Celsius=120 Celsius

Therefore,
Q=mc fita
=5x400x120
=240,000J
=240KJ


27000J of heat is required to change the temperature of 0.3kg of aluminum from 300 Celsius to 400.Find the specific heat capacity of aluminum and the heat capacity of the sample of aluminum.

Using,
Q=mc fita
Given that,
Q=27000J
m=0.3kg
fita=400-300 Celsius=100 Celcius

Q=mc fita
therefore to find c,use formula
c=Q/m x fita (note:to find either c,m,fita or Q,you must just do some cross multiplication)
=27000/(0.3x100)
=900J/kg/k


Using, C=mc
=0.3x900
=270J/k
 
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3000 ?
Lol !
I paid 2000/month at my academy in Lahore....all pastpapers from 1994-2010 in 1.5 months and got an A* !
 
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