Answer: The energy of combustion of butter is 31.5 kJ/g
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=C\times \Delta T[/tex]
Q = Heat absorbed by calorimeter =?
C = heat capacity of calorimeter = [tex]2.67kJ/^0C[/tex]
Initial temperature of the calorimeter = [tex]T_i[/tex] = [tex]23.5^0C[/tex]
Final temperature of the calorimeter = [tex]T_f[/tex] = [tex]26.1^0C[/tex]
Change in temperature ,[tex]\Delta T=T_f-T_i=(26.1-23.5)^0C=2.6^0C[/tex]
Putting in the values, we get:
[tex]Q=2.67kJ/^0C\times 2.6^0C=6.94kJ[/tex]
As heat absorbed by calorimeter is equal to heat released by combustion of butter
[tex]Q=q[/tex]
Heat released by 0.22 g of butter = 6.94 kJ
Heat released by 1g of butter = [tex]\frac{6.94}{0.22}\times 1=31.5kJ[/tex]
The energy of combustion of butter is 31.5 kJ/g
