Specific heat and Heat capacity
We learned from my previous post that system can either absorb heat from the surroundings or release heat to the surroundings. The emission and absorption of heat causes an object to change the temperature. The temperature change of an object when it absorbs or releases heat is determined by heat capacity. Heat capacity (C) of a substance is the amount of heat required to raise the temperature of a given quantity of a substance by 1 K (1oC). The greater the heat capacity, the greater the heat required to produce a given rise in temperature. The heat capacity of 1 mol of a substance is called its molar heat capacity. The heat capacity of 1 g of a substance is called specific heat capacity. Specific heat (s) of a substance is the amount of heat required to raise the temperature of 1 g of the substance by 1 degree Celsius. Specific heat is an intensive property, while heat capacity is an extensive property. The relationship between the heat capacity and specific heat is:
C = ms
where C is the heat capacity, s is the specific heat and m is the mass of the substance in grams. For example, the specific heat of water (s) is 4.184 J/g.oC, what is the heat capacity of 60 g of water?
C = 60 g (4.184 J/g.oC) = 251 J/oC
The unit of heat capacity is J/oC since g is cancelled out.
The specific heat of a substance can be determined by measuring the temperature change, ∆T, of a given amount of substance in grams , when it gains or loses a specific quantity of heat, q.
Specific heat = q/ m x ∆T
For example, 209 J is required to increase the temperature of 50 g water by 1 K . What is the specific heat of water?
Specific Heat = 209 J / 50 g (1.00 K) = 4.18 J/g.K
If the specific heat and the amount of substance is given so as the change in temperature, the heat that is absorbed or released can be calculated. Look at the equation below:
q = ms∆T
q = C∆T
∆T is the change in temperature. ∆T = Tfinal - Tinitial. For an endothermic process q is also positive and for exothermic process the q is negative.
For example:
A 466 g of water is heated from 8.50oC to 74.60oC. Calculate the amount of heat absorbed in kJ by the water.
For example:
A 466 g of water is heated from 8.50oC to 74.60oC. Calculate the amount of heat absorbed in kJ by the water.
Solution:
q = ms∆T
q = 466 g (4.184 J/g.oC) (74.60oC - 8.50oC)
= 1949.7 oC x 66.1 oC
= 128875 J x( 1kJ/1000 J)
= 128875 J x( 1kJ/1000 J)
= 129 kJ
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