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Em um calorímetro de capacidade térmica 30 cal/º C e temperatura 40º C, são colocados um bloco maciço de uma liga metálica desconhecida de 200 g que estava a 120º C de temperatura. Junto ao calorímetro foram colocados 100g de água na mesma temperatura do calorímetro. Sabendo que após certo tempo atinge-se o equilíbrio térmico, encontre a temperatura aproximada no qual acontece esse equilíbrio. Dados: calor específico da água = 1 cal/g ºC e calor específico da liga = 03, cal/g ºC

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Resposta:

Let's break down this thermometry problem step by step!

We have a calorimeter with a thermal capacity of 30 cal/°C, initially at a temperature of 40°C. A 200g metal block of unknown metal, initially at 120°C, is placed in the calorimeter. Alongside the calorimeter, 100g of water at the same temperature as the calorimeter (40°C) is added. After some time, thermal equilibrium is reached. We need to find the approximate temperature at which this equilibrium occurs.

First, let's identify the unknowns and the given data:

  • mmetal = 200g (mass of the metal block)
  • cmetal = 0.3 cal/g°C (specific heat capacity of the metal)
  • mwater = 100g (mass of water)
  • cwater = 1 cal/g°C (specific heat capacity of water)
  • Tinitial metal = 120°C (initial temperature of the metal block)
  • Tinitial calorimeter = 40°C (initial temperature of the calorimeter)
  • Ccalorimeter = 30 cal/°C (thermal capacity of the calorimeter)

When the system reaches thermal equilibrium, the temperature of the metal block, water, and calorimeter will be the same. Let's call this temperature Tf.

We can use the principle of heat transfer to set up an equation. The heat gained by the calorimeter and water is equal to the heat lost by the metal block:

$$Q_{gained} = Q_{lost}$$

Breaking it down further:

$$C_{calorimeter} times (T_f - T_{initial calorimeter}) + m_{water} times c_{water} times (T_f - T_{initial calorimeter}) = m_{metal} times c_{metal} times (T_{initial metal} - T_f)$$

Simplifying the equation:

$$30 times (T_f - 40) + 100 times 1 times (T_f - 40) = 200 times 0.3 times (120 - T_f)$$

Solving for Tf:

$$T_f = frac{30 times 40 + 100 times 40 + 200 times 0.3 times 120}{30 + 100 + 200 times 0.3} = 65°C$$

Therefore, the correct answer is A) 65°C.

Explanation: During the heat transfer process, the metal block loses heat, while the calorimeter and water gain heat. By applying the principle of heat transfer and using the given data, we can set up an equation to find the final temperature at which thermal equilibrium occurs. Solving the equation, we get Tf = 65°C, which is the correct answer.

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