A container contains 200 mL of 100% proof alcohol i.e., it has 50% ethyl alcohol and 50% water by volume) at 20°C. How much heat is needed to bring the mixture to the boiling point of the alcohol? Assume that the specific heat in the 100% proof can be treated as due to the alcohol and water separately, and the density, boiling point, and specific heat of alcohol are 0.81 g/cm³, 78°C, and 2.4 J/gC°), respectively.)

Question

Quizplus · Accepted Answer

First, we need to find out the mass of the alcohol in the container. Using the density of alcohol, we can convert the volume of the solution to its mass as follows:

Density of alcohol = 0.81 g/cm³
Volume of solution = 200 mL = 0.2 L
Mass of solution = Density x Volume = 0.81 x 0.2 = 0.162 kg

Since the solution is a 50-50 mixture of alcohol and water, half of the mass is alcohol and the other half is water. Therefore, the mass of alcohol is 0.081 kg.

Next, we need to find the amount of heat required to raise the temperature of the alcohol from 20°C to its boiling point of 78°C. We can use the specific heat of alcohol to do this:

Specific heat of alcohol = 2.4 J/g°C
Change in temperature = 78°C - 20°C = 58°C
Amount of heat required = Mass x Specific heat x Change in temperature
= 0.081 x 2.4 x 58
= 11.136 J

Therefore, the amount of heat needed to bring the mixture to the boiling point of alcohol is 11,136 J. However, since the question asks for the total heat needed (i.e., including any latent heat of vaporization), we need to add the heat required for vaporization of the alcohol. This can be calculated using the heat of vaporization of alcohol, which is 846 J/g:

Heat of vaporization of alcohol = 846 J/g
Amount of heat required for vaporization = Mass x Heat of vaporization
= 0.081 x 846
= 68.526 J

Therefore, the total amount of heat needed is:

Total heat = 11,136 + 68.526
= 35,525 J

Hence, the best choice is B.

A Container Contains 200 ML of 100% Proof Alcohol I