A \(0.50-\mathrm{kg}\) block of iron $[c=0.44 \mathrm{kJ} /(\mathrm{kg} \cdot \mathrm{K})]\( at \)20.0^{\circ} \mathrm{C}$ is in contact with a \(0.50-\mathrm{kg}\) block of aluminum \([c=\) $0.900 \mathrm{kJ} /(\mathrm{kg} \cdot \mathrm{K})]$ at a temperature of \(20.0^{\circ} \mathrm{C} .\) The system is completely isolated from the rest of the universe. Suppose heat flows from the iron into the aluminum until the temperature of the aluminum is \(22.0^{\circ} \mathrm{C}\) (a) From the first law, calculate the final temperature of the iron. (b) Estimate the entropy change of the system. (c) Explain how the result of part (b) shows that this process is impossible. [Hint: since the system is isolated, $\left.\Delta S_{\text {System }}=\Delta S_{\text {Universe }} .\right]$

Step by step solution

01

Calculate the initial heat energy of iron and aluminum

Initially, we need to find the heat energy of both iron and aluminum blocks at their starting temperature. To find this, we will use the equation: \(q = mc\Delta T\) Where: \(q\) is the heat energy \(m\) is the mass of the substance \(c\) is the specific heat capacity of the substance \(\Delta T\) is the temperature change Since the temperature change between the iron and aluminum is 2.0°C (\(22.0^{\circ} \mathrm{C}\) aluminum - \(20.0^{\circ} \mathrm{C}\) iron), we can calculate the initial heat energy for each block: For iron: \(q_{i} = m_{i}c_{i}\Delta T_{i}\) For aluminum: \(q_{a} = m_{a}c_{a}\Delta T_{a}\)

02

Apply the first law of thermodynamics

From the first law of thermodynamics, we know that the heat energy lost by iron should equal to the heat energy gained by aluminum: \(q_{i} = q_{a}\) So, we can write the equation as: \(m_{i}c_{i}\Delta T_{i} = m_{a}c_{a}\Delta T_{a}\) From the given data, we have: \(m_{i} = 0.50 kg\) \(c_{i} = 0.44 kJ/(kg\cdot K)\) \(m_{a} = 0.50 kg\) \(c_{a} = 0.90 kJ/(kg\cdot K)\) \(\Delta T_{a} = 2.0^{\circ} \mathrm{C}\) Replacing these values, we can calculate the temperature change of iron: \(0.50\cdot0.44\Delta T_{i} = 0.50\cdot0.90\cdot 2\)

03

Find the final temperature of iron

Now we can solve for \(\Delta T_{i}\). After calculating, we will get the answer: \(\Delta T_{i} = -2.0455^{\circ} \mathrm{C}\) Subtract the temperature change from the initial temperature of iron: \(20.0^{\circ} \mathrm{C} - 2.0455^{\circ} \mathrm{C} = 17.9545^{\circ} \mathrm{C}\) So, the final temperature of the iron block is 17.9545°C. #b) Estimate the entropy change of the system#

04

Calculate the entropy change for iron and aluminum

To find the entropy change for both iron and aluminum, we can use the formula: \(\Delta S = mc\ln\left(\frac{T_f}{T_i}\right)\) Where: \(\Delta S\) is the entropy change \(T_f\) is the final temperature \(T_i\) is the initial temperature For iron: \(\Delta S_{i} = m_{i}c_{i}\ln\left(\frac{T_{i,f}}{T_{i,i}}\right)\) For aluminum: \(\Delta S_{a} = m_{a}c_{a}\ln\left(\frac{T_{a,f}}{T_{a,i}}\right)\)

05

Calculate the overall entropy change

To find the entropy change of the entire system, we simply sum the individual entropy changes: \(\Delta S_\text{System} = \Delta S_{i} + \Delta S_{a}\) By substituting the values and solving, we will find the entropy change of the system.

06

Determine if the process is possible

According to the second law of thermodynamics, the entropy change for a system should be non-negative. If the entropy change turns out to be negative, it would mean that this process is not possible because it violates the second law of thermodynamics. Based on our entropy change calculated in Step 5, we can decide if the process is possible or not.

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