Question

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.

Answer 1

Answer:

3984.12 J joules of heat are required to heat 25.0 g of isopropyl alcohol to its boiling point.

Explanation:

$Q=m\times c\Delta T=m\times c\times (T_2-T_1)$

Where:

Q = heat absorbed or heat lost

c = specific heat of substance

m = Mass of the substance  

ΔT = change in temperature of the substance

$T_1$ = Initial temperature of the substance

$T_2$ = 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 = $T_1=21.2^oC$

Final temperature of the isopropyl alcohol = $T_2=82.4 ^oC$

Heat absorbed by the isopropyl alcohol to boil:

$Q=25.0 g\times 2.604 J/g^oC\times (82.4^oC-21.2^oC)=3984.12 J$

3984.12 J joules of heat are required to heat 25.0 g of isopropyl alcohol to its boiling point.