Question

A \(0.500-\mathrm{kg}\) block of iron at \(60.0^{\circ} \mathrm{C}\) is placed in contact with a \(0.500-\mathrm{kg}\) block of iron at $20.0^{\circ} \mathrm{C} .\( (a) The blocks soon come to a common temperature of \)40.0^{\circ} \mathrm{C} .$ Estimate the entropy change of the universe when this occurs. [Hint: Assume that all the heat flow occurs at an average temperature for each block.] (b) Estimate the entropy change of the universe if, instead, the temperature of the hotter block increased to \(80.0^{\circ} \mathrm{C}\) while the temperature of the colder block decreased to \(0.0^{\circ} \mathrm{C}\) [Hint: The answer is negative, indicating that the process is impossible. \(]\)

Short answer

Answer: The entropy change of the universe in this case is \(0 J/K\). However, it is important to note that this process violates the second law of thermodynamics, as heat would need to flow from colder to hotter objects, which is impossible in practice.

Step-by-step solution

Part (a): Entropy change for both blocks coming to a common temperature of \(40.0^{\circ} \mathrm{C}\)

First, we need to calculate the entropy change for each block. For this, we will use the formula for entropy change given by \(\Delta S = mc \ln{\frac{T_{f}}{T_{i}}}\). Let's consider the first block (Hot Block), with an initial temperature of \(60^{\circ}C\) and final temperature of \(40^{\circ}C\). The mass of the block is \(0.500 kg\) and the specific heat capacity of iron is \(450 \frac{J}{kg∙K}\). So, the entropy change for the first block is: \(\Delta S_{1} = (0.500 kg)(450 \frac{J}{kg∙K})\ln{\frac{313 K}{333 K}} = -63.7 J/K\) Now, let's calculate the entropy change for the second block (Cold Block) with an initial temperature of \(20^{\circ}C\) and final temperature of \(40^{\circ}C\): \(\Delta S_{2} = (0.500 kg)(450 \frac{J}{kg∙K})\ln{\frac{313 K}{293 K}} = 63.7 J/K\)

Total entropy change for part (a)

Now, let's calculate the total entropy change in the universe by adding the entropy changes of both blocks: \(\Delta S_{total} = \Delta S_{1} + \Delta S_{2} = -63.7 J/K + 63.7 J/K = 0 J/K\) In this case, the total entropy change of the universe is zero, which suggests that the process is reversible.

Part (b): Entropy change when hotter block increases to \(80.0^{\circ}C\) and colder block decreases to \(0.0^{\circ}C\)

Similar to part (a), we will calculate the entropy change for each block individually and then sum up the entropy changes to get the total entropy change of the universe. For the first block (Hot Block), with an initial temperature of \(60^{\circ}C\) and final temperature of \(80^{\circ}C\): \(\Delta S_{1} = (0.500 kg)(450 \frac{J}{kg∙K})\ln{\frac{353 K}{333 K}} = 31.9 J/K\) For the second block (Cold Block), with an initial temperature of \(20^{\circ}C\) and final temperature of \(0^{\circ}C\): \(\Delta S_{2} = (0.500 kg)(450 \frac{J}{kg∙K})\ln{\frac{273 K}{293 K}} = -31.9 J/K\)

Total entropy change for part (b)

Now, let's calculate the total entropy change in the universe by adding the entropy changes of both blocks: \(\Delta S_{total} = \Delta S_{1} + \Delta S_{2} = 31.9 J/K - 31.9 J/K = 0 J/K\) In this case, the total entropy change of the universe is also zero, which indicates that the process is reversible. However, based on the initial conditions, this process violates the second law of thermodynamics, as heat would need to flow from colder to hotter objects, which is impossible in practice.