Dentro de um calorímetro ideal de capacidade térmica desprezível são colocados 200 gramas de água líquida a 20 ºC , 90 gramas de gelo a − 10 ºC e 10 gramas de vapor de água a 105 ºC. Considere os seguintes dados: Calores específicos da água, do gelo e do vapor, respectivamente iguais a 1 cal / (g ºC), 0,5 cal / (g ºC) e 0,5 cal / (g ºC); Calores latentes de fusão e de vaporização da água, respectivamente iguais a 80 cal / g e 540 cal / g . Supondo que o sistema não troque calor com o calorímetro e com o ambiente externo, podemos afirmar que, após o equilíbrio térmico a temperatura final do sistema será de

Let's break down the problem step by step:

First, we need to find the total heat required to bring the system to thermal equilibrium. We can do this by calculating the heat required to melt the ice, heat the water, and condense the steam.

For the ice, we need to calculate the heat required to melt it. Since the temperature of the ice is -10°C, we need to heat it to 0°C to melt it. The specific heat of ice is 0.5 cal/(g°C), so the heat required is:

Q_{ice} = m_{ice} cdot c_{ice} cdot Delta T = 90 g cdot 0.5 cal/(g°C) cdot (0°C - (-10°C)) = 450 cal

Once the ice is melted, we need to heat the resulting water to the final temperature. The specific heat of water is 1 cal/(g°C), so the heat required is:

Q_{water} = m_{water} cdot c_{water} cdot Delta T = (200 g + 90 g) cdot 1 cal/(g°C) cdot Delta T

Now, we need to condense the steam. The specific heat of steam is 0.5 cal/(g°C), and the heat required is:

Q_{steam} = m_{steam} cdot c_{steam} cdot Delta T = 10 g cdot 0.5 cal/(g°C) cdot Delta T

We also need to take into account the latent heat of fusion and vaporization. The latent heat of fusion is 80 cal/g, and the latent heat of vaporization is 540 cal/g. So, the total heat required is:

Q_{total} = Q_{ice} + Q_{water} + Q_{steam} + m_{ice} cdot L_f + m_{steam} cdot L_v = 450 cal + (200 g + 90 g) cdot 1 cal/(g°C) cdot Delta T + 10 g cdot 0.5 cal/(g°C) cdot Delta T + 90 g cdot 80 cal/g + 10 g cdot 540 cal/g

Since the system is in thermal equilibrium, the total heat required is zero. So, we can set up the equation:

450 cal + (290 g) cdot 1 cal/(g°C) cdot Delta T + 10 g cdot 0.5 cal/(g°C) cdot Delta T + 7200 cal + 5400 cal = 0

Simplifying the equation, we get:

290 g cdot Delta T + 5 g cdot Delta T = -12650 cal

Solving for ΔT, we get:

Delta T = -9.25°C

Since the initial temperature of the water is 20°C, the final temperature is:

T_f = 20°C - 9.25°C = 9.25°C

Therefore, the correct answer is E) 9.25°C.

This problem requires careful calculation of the heat required for each process and taking into account the latent heat of fusion and vaporization. The key is to set up the correct equation and simplify it to find the final temperature.