Suppose that 1.01g of rubbing alcohol(C3H8O) evaporates from a 56.0g aluminum block. If the aluminum block is initially at 25'C, what is the final temperature of the block after the evaporation of the alcohol? Assume that the heat required for the vaporization of the alcohol comes only from the aluminum block and that the alcohol vaporizes at 25'C

1. Look up the heat of vaporization of isopropyl alcohol. The figure I found is 44.0 kJ/mole. Divide that by the molar mass to get the kJ/g. Multiply by 1.01g to get the heat absorbed from the aluminum block.

2. The applicable formula for the aluminum is:
heat = (specific heat of Al)(grams Al)(deltaT). Look up the specific heat of Al. Solve for deltaT, the temp. change. Subtract that from 25.0ºC.

15.1

To calculate the final temperature of the aluminum block after the evaporation of rubbing alcohol, we can use the concept of heat transfer through the equation:

Q = m × c × ∆T

Where:
Q: Heat transfer (in Joules)
m: Mass of the substance (in grams)
c: Specific heat capacity (in J/g°C)
∆T: Change in temperature (in °C)

In this case, we need to consider the heat transferred from the aluminum block to the alcohol to vaporize it. We first calculate the heat required to vaporize the rubbing alcohol:

Q_alcohol = m_alcohol × ∆H_vaporization

Where:
Q_alcohol: Heat required to vaporize alcohol (in Joules)
m_alcohol: Mass of rubbing alcohol evaporated (in grams)
∆H_vaporization: Heat of vaporization of rubbing alcohol (in J/g)

Using the given values:
m_alcohol = 1.01g
∆H_vaporization = the heat of vaporization of rubbing alcohol (not provided in the question)

We need the heat of vaporization (∆H_vaporization) value for rubbing alcohol to proceed. This value is typically available in reference books or online databases. Let's assume it is 202 J/g for rubbing alcohol.

Substituting the known values:
Q_alcohol = 1.01g × 202 J/g

Next, we calculate the heat transferred from the aluminum block to the alcohol:

Q_aluminum = m_aluminum × c_aluminum × ∆T

Where:
Q_aluminum: Heat transferred from the aluminum block (in Joules)
m_aluminum: Mass of the aluminum block (in grams)
c_aluminum: Specific heat capacity of aluminum (in J/g°C)
∆T: Change in temperature of the aluminum block (the final temperature - initial temperature)

Substituting the known values:
m_aluminum = 56.0g
c_aluminum = the specific heat capacity of aluminum (not provided in the question)

We need the specific heat capacity (c_aluminum) value for aluminum to proceed. The specific heat capacity of aluminum is approximately 0.897 J/g°C.

Substituting the values in the equation:
Q_aluminum = 56.0g × 0.897 J/g°C × ∆T

Since the heat transfer from the aluminum block is equal to the heat required to vaporize the alcohol, we can equate the two equations:

Q_aluminum = Q_alcohol
56.0g × 0.897 J/g°C × ∆T = 1.01g × 202 J/g

Now, solve for ∆T to find the change in temperature of the aluminum block:

56.0g × 0.897 J/g°C × ∆T = 1.01g × 202 J/g
∆T = (1.01g × 202 J/g) / (56.0g × 0.897 J/g°C)

∆T ≈ 3.61°C

To find the final temperature, we add the change in temperature (∆T) to the initial temperature:

Final temperature = Initial temperature + ∆T
Final temperature = 25°C + 3.61°C

The final temperature of the aluminum block after the evaporation of rubbing alcohol is approximately 28.61°C.