Um calorímetro adiabático de alumínio, com 300g de massa, contém 400g de água a 22°C.
Uma amostra de 100g de gelo a -20ºC é introduzida no calorímetro. Com isso, verifica-se que o sistema atinge uma temperatura T no equilíbrio térmico. Com o sistema já em equilíbrio térmico, uma segunda amostra de gelo, com 100g e temperatura de -10ºC, é adicionada e verifica-se que um novo equilíbrio térmico é atingido quando ainda resta certa massa M de gelo não derretida.
Considere os seguintes dados:
Calor específico do gelo = 0,5 cal/g.°C
Calor específico da água = 1,0 cal/g.°C
Calor específico do alumínio = 0,22 cal/g.°C
Calor latente de fusão do gelo = 80 cal/g
Calor latente de vaporização da água = 540 cal/g
De acordo com os dados acima, a temperatura T e a massa M são, aproximadamente
- A) T = 2,21°C e M = 9,40g
- B) T = 2,21°C e M = 90,6g
- C) T = 0°C e M = 9,40g
- D) T = 3,98°C M = 90,6g
Resposta:
To solve this problem,<|begin_of_text|>201C we need to use the principle of heat transfer and the concept of heat capacity. Let's break down the process step by step.
Initially, we have 400g of water at 22°C and a calorimeter with a mass of 300g made of aluminum. The calorimeter contains 400g of water.
When the 100g of ice at -20°C is introduced into the calorimeter, the system reaches a thermal equilibrium temperature T. To find this temperature, we need to consider the heat transfer that occurs between the calorimeter, the water, and the ice.
The heat required to melt the ice is given by:$$Q_1 = mL_f = 100g times 80 dfrac{cal}{g} = 8000 cal$$
The heat absorbed by the calorimeter is given by:$$Q_2 = mcDelta T = 300g times 0.22 dfrac{cal}{g°C} times (T - 22°C)$$
The heat absorbed by the water is given by:$$Q_3 = mcDelta T = 400g times 1.0 dfrac{cal}{g°C} times (T - 22°C)$$
Since the heat gained by the calorimeter and the water is equal to the heat lost by the ice, we can set up the following equation:$$Q_1 = Q_2 + Q_3$$
Substituting the values, we get:$$8000 cal = 300g times 0.22 dfrac{cal}{g°C} times (T - 22°C) + 400g times 1.0 dfrac{cal}{g°C} times (T - 22°C)$$
Solving for T, we get:$$T = 2.21°C$$
Now, when the second sample of ice at -10°C is added, the system reaches a new thermal equilibrium temperature. Let's assume that x grams of ice remain unmelted.
The heat gained by the calorimeter and the water is given by:$$Q_4 = mcDelta T = (300g + 400g) times 0.22 dfrac{cal}{g°C} times (T - 2.21°C)$$
The heat gained by the melting of the ice is given by:$$Q_5 = mL_f = xg times 80 dfrac{cal}{g}$$
Since the heat gained by the calorimeter and the water is equal to the heat lost by the ice, we can set up the following equation:$$Q_4 = Q_5$$
Substituting the values, we get:$$(300g + 400g) times 0.22 dfrac{cal}{g°C} times (T - 2.21°C) = xg times 80 dfrac{cal}{g}$$
Solving for x, we get:$$x = 90.6g$$
Therefore, the correct answer is B) T = 2.21°C and M = 90.6g.
Explanation: The key to solving this problem is to identify the heat transfer processes that occur between the calorimeter, the water, and the ice. By using the principles of heat transfer and the concept of heat capacity, we can set up equations that relate the heat gained by the calorimeter and the water to the heat lost by the ice. By solving these equations, we can find the thermal equilibrium temperature T and the mass M of the remaining unmelted ice.
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