Answer:
The heat required to heat 25.0 g of isopropyl alcohol to its boiling point is 3984.12 Joules.
Explanation:
Mass of isopropyl alcohol = m = 25.0 g
Specific heat of isopropyl alcohol  = c = 2.604 J/g°C
Initial temperature of the isopropyl alcohol = [tex]T_1=21.2^oC[/tex]
Final temperature of the isopropyl alcohol = [tex]T_2=82.4 ^oC[/tex]
Heat absorbed by the isopropyl alcohol to boil: Q
[tex]Q=m\times c\times \Delta T=m\times c\times (T_2-T_1)[/tex]
Where:
Q = heat absorbed or heat lost
c = specific heat of substance
m = Mass of the substance Â
ΔT = change in temperature of the substance
[tex]T_1[/tex] = Initial temperature of the substance
[tex]T_2[/tex] = Final temperature of the substance
On substituting the values  :
[tex]Q=25.0 g\times 2.604 J/g^oC\times (82.4^oC-21.2^oC)=3984.12 J[/tex]
The heat required to heat 25.0 g of isopropyl alcohol to its boiling point is 3984.12 Joules.
