Question

20. Suppose that 1.02 g of rubbing alcohol (C3H8O) evaporates from a 55.0 g block of aluminum that is initially at 25 °C.

What is the final temperature of the aluminum?

The ∆Hvap for the alcohol is 45.4 kJ/mol at 25 °C.

Assume that the heat required for the vaporization of alcohol comes only from the aluminum block and the alcohol vaporizes at 25 °C. The specific heat capacity of Al is 0.903 J/g°C.

Answer 1

Ans. Given, Mass of alcohol = 1.02 g

Now,

            Moles of alcohol = Mass/ molar mass

                                                = 1.02 g / (60.09592 g/ mol)

                                                = 0.01697 mol

# Given, dHvap of alcohol = 45.4 kJ/ mol at 25.0⁰C

Now,

Total heat absorbed by alcohol for evaporation = dHvap x moles of alcohol evaporated

                                                = (45.4 kJ/ mol) x 0.01697 mol

                                                = 0.770438 kJ

                                                = 770.438 J

# Since alcohol absorbs heat from Al block, the total amount of heat absorbed by alcohol must be equal to the total heat lost be Al block.

So,

            Heat lost by Al = - 770.438 J         ; [the –ve sign indicates heat loss]

# The required Heat loss is given by-

q = m s dT                            - equation 1

Where,

q = heat lost

m = mass

s = specific heat

dT = Final temperature – Initial temperature

Let the final temperature be T

Putting the values in equation 1-

            -770.438 J = 55.0 g x (0.903 J g^-10C^-1) x (T – 25.0⁰C)

            Or, -770.438 J / (49.665 J ⁰C^-1) = (T – 25.0⁰C)

            Or, -15.51⁰C = T – 25.0⁰C

            Or, T = -15.51⁰C + 25.0⁰C

            Hence, T = 9.49⁰C

Therefore, final temperature of Al block = T = 9.49⁰C