Answer:
3984.12 J joules of heat are required to heat 25.0 g of isopropyl alcohol to its boiling point.
Explanation:
[tex]Q=m\times c\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
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
We have 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:
[tex]Q=25.0 g\times 2.604 J/g^oC\times (82.4^oC-21.2^oC)=3984.12 J[/tex]
3984.12 J joules of heat are required to heat 25.0 g of isopropyl alcohol to its boiling point.
