Question

An iron block of unknown mass at \(85^{\circ} \mathrm{C}\) is dropped into an insulated tank that contains \(100 \mathrm{L}\) of water at \(20^{\circ} \mathrm{C}\). At the same time, a paddle wheel driven by a 200 -W motor is activated to stir the water. It is observed that thermal equilibrium is established after 20 min with a final temperature of \(24^{\circ} \mathrm{C} .\) Assuming the surroundings to be at \(20^{\circ} \mathrm{C}\), determine (a) the mass of the iron block and ( \(b\) ) the exergy destroyed during this process. Answers: (a) \(52.0 \mathrm{kg},\) (b) \(375 \mathrm{kJ}\)

Short answer

The mass of the iron block is 52 kg. b) Calculate the exergy destroyed during the process. The exergy destroyed during the process is 375 kJ.

Step-by-step solution

Energy balance equation

First, we will use the energy balance equation, which states that the energy input to the system must equal the sum of the energy stored within the system and the energy leaving the system. We will write the equation: \(Q + W = m_i c_i (T_f - T_i)\), where \(Q\) is the heat added to the system, \(W\) is the work done by the paddle wheel, \(m_i\) is the mass of the iron block, \(c_i\) is the specific heat capacity of iron, \(T_f\) is the final temperature, and \(T_i\) is the initial temperature of the iron block.

Calculate the work done by the paddle wheel

Given that the paddle wheel is driven by a 200 W motor for 20 minutes, we can calculate the work done by the paddle wheel as: \(W = P \times t = (200\,\text{W})\times (20\,\text{min}\times 60\,\text{s/min}) = 240000\, \text{J}\)

Mass balance equation

We will now write the mass balance equation to relate the masses of the iron block and the water: \(m_i = m_w + m_b\), where \(m_w\) is the mass of the water and \(m_b\) is the mass of the iron block.

Calculate the mass of water

The volume of the water is given as 100 L, and we can use the density of water to calculate the mass: \(m_w = \rho V = (1000\, \text{kg/m}^3)(100\, \text{L} \times 0.001\, \text{m}^3/\text{L}) = 100\, \text{kg}\)

Apply specific heat capacities

In this step, we will apply the specific heat capacities of water and iron. Given that: Iron: \(c_i = 442\, \text{J/kg} \cdot \text{K}\), Water: \(c_w = 4186\, \text{J/kg} \cdot \text{K}\)

Equate the energy balance equation

Equate the energy balance equation and solve for the mass of the iron block, \(m_b\): \(Q + W = m_i c_i(T_f - T_i) + m_w c_w ( T_f - T_w)\) Since the system is insulated, \(Q=0\). Plug in the values for \(T_f, T_w, W, c_i, c_w, m_w\) and solve for \(m_b\): \(240000\, \text{J} = m_b(442\, \text{J/kg} \cdot \text{K})(24\,^\circ \mathrm{C} - 85\,^\circ \mathrm{C}) + (100\, \text{kg})(4186\, \text{J/ kg} \cdot \text{K})(24\,^\circ \mathrm{C} - 20\,^\circ \mathrm{C})\) Solve for \(m_b\): \(m_b = 52\, \text{kg}\) (Answer a)

Calculate the exergy destroyed

In this step, we will calculate the exergy destroyed using the second law of thermodynamics. The exergy destruction can be calculated as: \(\Delta E_{x,dest} = \int_{t_1}^{t_2} \left(T_0 \frac{dS}{dt} - \frac{dE}{dt}\right) dt\) Where \(T_0\) is the surrounding temperature, \(dS/dt\) is the rate of entropy generation, and \(dE/dt\) is the rate of energy transfer. Using the values from the energy balance and mass of the iron block, we can calculate the exergy destroyed: \(\Delta E_{x,dest} = T_0 \Delta S_{gen} - E_{trans}\) The entropy generation can be calculated as \(\Delta S_{gen} = m_b c_i \log \frac{T_f}{T_i} + m_w c_w \log \frac{T_f}{T_w}\) , and \(E_{trans}=W\). Plugging in the values, we calculate the exergy destruction as: \(\Delta E_{x,dest} = (20^\circ \mathrm{C} + 273.15\, \text{K})((52\, \text{kg})(442\, \text{J/kg} \cdot \text{K})\log\frac{297.15\, \text{K}}{358.15\, \text{K}} + (100\, \text{kg})(4186\, \text{J/kg} \cdot \text{K})\log\frac{297.15\, \text{K}}{293.15\, \text{K}}) - 240000\, \text{J}\) \(\Delta E_{x,dest} = 375\, \text{kJ}\) (Answer b)