Answer : The correct option is, [tex]30.9^oC[/tex]
Explanation :
Formula used :
[tex]q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
where,
[tex]q[/tex] = heat released = 24 KJ
[tex]m[/tex] = mass of bomb calorimeter = 1.30 Kg
[tex]c[/tex] = specific heat = [tex]3.41J/g^oC[/tex]
[tex]T_{final}[/tex] = final temperature = ?
[tex]T_{initial}[/tex] = initial temperature = [tex]25.5^oC[/tex]
Now put all the given values in the above formula, we get the final temperature of the calorimeter.
[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]24KJ=1.30Kg\times 3.41J/g^oC\times (T_{final}-25.5)^oC[/tex]
[tex]T_{final}=30.9^oC[/tex]
Therefore, the final temperature of the calorimeter is, [tex]30.9^oC[/tex]
Answer:
Final temperature is 30,9°C
Explanation:
A bomb calorimeter is an instrument that allows the determination of the heat of combustion of a reaction. The heat q is defined as:
q = C×m×ΔT (1)
Where:
C is specif heat of calorimeter (3,41 J/g°C)
m is mass (1,3kg ≡ 1300g)
And ΔT is final temperature - initial temperature (X-25,5°C) X is final temperature
And q is heat (24,0kJ ≡ 24000J)
Replacing in (1):
24000J = 3,41J/g°C×1300g×(X-25,5°C)
5,41 = X-25,5°C
25,5°C + 5,41 = X
30,91°C = X
Final temperature is 30,9°C
I hope it helps!
