How many joules of heat are required to heat 25.0 g of isopropyl alcohol from the prevailing room temperature, 21.2 oC, to its boiling point, 82.4 oC? The specific heat of isopropyl alcohol is 2.604 J/g °C.

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 =

Final temperature of the isopropyl alcohol =

Heat absorbed by the isopropyl alcohol to boil: Q

Where:

Q = heat absorbed or heat lost

c = specific heat of substance

m = Mass of the substance  

ΔT = change in temperature of the substance

= Initial temperature of the substance

= Final temperature of the substance

On substituting the values  :

The heat required to heat 25.0 g of isopropyl alcohol to its boiling point is 3984.12 Joules.

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