Uma massa de 0,15 kg de vapor d’água a 120 oC perde 108000 cal para o ambiente. Considerando o calor específico do vapor como 0,5 cal/goC, o calor específico da água igual a 1 cal/gºC, o calor latente de condensação da água igual a -540 cal/g e o calor latente de solidificação do gelo igual a -80 cal/g, teremos ao final do processo

Let's break down the problem step by step:

The initial situation is a 0.15 kg vapor of water at 120°C, which loses 10800 cal to the environment. To find the final state of the system, we need to analyze the possible scenarios.

First, let's calculate the temperature change of the vapor. Since the heat lost is 10800 cal, and the specific heat of vapor is 0.5 cal/g°C, we can calculate the temperature change:

ΔQ = mcΔT => 10800 cal = 0.15 kg × 0.5 cal/g°C × ΔT

ΔT = 10800 cal / (0.15 kg × 0.5 cal/g°C) = 144°C

Since the initial temperature is 120°C, the final temperature of the vapor would be:

Tf = 120°C - 144°C = -24°C

However, this is not possible, as the temperature cannot be lower than 0°C. Therefore, the vapor must have undergone a phase change.

Let's consider the possibilities:

A) 150 g of water at 2.7°C: This option is not possible, as the temperature is not consistent with the calculation above.

B) 150 g of ice at 0°C: This option is also not possible, as the temperature change would require more heat to be lost.

C) 131.25 g of ice and 18.75 g of liquid water: This option is the most plausible, as it would require a phase change from vapor to liquid and then to ice, which is consistent with the heat lost.

D) 18.75 g of ice and 131.25 g of liquid water: This option is similar to C, but with the masses reversed, which is not possible given the initial conditions.

Therefore, the correct answer is C) 131.25 g of ice and 18.75 g of liquid water.

Explanation:

The key to this problem is to recognize that the heat lost is not sufficient to change the temperature of the vapor by 144°C. Therefore, a phase change must have occurred. By analyzing the possible scenarios, we conclude that the most plausible option is C) 131.25 g of ice and 18.75 g of liquid water, which is consistent with the heat lost and the initial conditions.