Let's analyze and solve the problem step-by-step. --- # Problem Breakdown - **Given Data:** - **Liquid metal (L)** is poured into a mold. - Temperature of mold **\(T_p\)** at pouring: **25°C** - **Cooling process:** When mold reaches **thermal equilibrium** instantly, the temperature drops to **\(T_s\)** (freezing temperature of the liquid). - Properties: - **Thermal conductivity, \(k\)** =1.6 W/m·K - **Density, \(\rho\)** = 8.52 × 100 kg/m³ - **Specific heat capacity, \(C_p\)** = 100 J/kg·K - **Specific heat capacity of mold, \(C_{p,mould}\)** = 385 J/kg·K - **Heat of fusion, \(L_f\)** = 3.2 × 100 kJ/m³ --- # Step 1: Understand the physical process - The liquid metal cools from \(T_p = 25^\circ C\) to the freezing temperature \(T_f\) (assumed to be °C unless specified otherwise). - The process is **instantaneous** at the moment the mold reaches thermal equilibrium. - The heat lost by the liquid metal is used to: 1. Cool the liquid metal to \(T_f\) 2. Freeze the liquid (latent heat) --- # Step 2: Write the energy balance The heat lost by the liquid metal during cooling and freezing: \[ Q_{metal} = \text{Cooling} + \text{Freezing} \] - Cooling from \(25^\circ C\) to \(^\circ C\): \[ Q_{cool} = m \times C_p \times (T_p - T_f) \] - Freezing (latent heat): \[ Q_{latent} = m \times L_f \] Total heat: \[ Q_{total} = m \times C_p \times (T_p - T_f) + m \times L_f \] --- # Step 3: Express mass in terms of volume Given the **volume of the liquid metal**: \[ V = 1 \text{ m}^3 \] Density: \[ \rho = 8.52 \times 10^3 \text{ kg/m}^3 \] Mass: \[ m = \rho \times V = 8.52 \times 10^3 \text{ kg} \] --- # Step 4: Calculate heat loss from the liquid metal \[ Q_{metal} = m \times C_p \times (T_p - T_f) + m \times L_f \] Plugging in the values: \[ Q_{metal} = (8.52 \times 10^3) \times 100 \times (25 - ) + (8.52 \times 10^3) \times 3.2 \times 10^6 \] Calculations: - Cooling: \[ 8.52 \times 10^3 \times 100 \times 25 = 8.52 \times 10^3 \times 25 \times 10^3 = (8.52 \times 25) \times 10^6 = 213 \times 10^6 \text{ J} \] - Freezing: \[ 8.52 \times 10^3 \times 3.2 \times 10^6 = (8.52 \times 3.2) \times 10^9 = 27.264 \times 10^9 \text{ J} \] Total heat: \[ Q_{total} = 213 \times 10^6 + 27.264 \times 10^9 \approx 27.477 \times 10^9 \text{ J} \] --- # Step 5: Heat transfer to the mold Assuming the mold's thermal capacity is negligible and that the heat transfer is **instantaneous** at thermal equilibrium, the heat extracted from the liquid metal is transferred to the mold, which is at a lower temperature. The heat transfer rate: \[ Q = h \times A \times (T_{liquid} - T_{mould}) \] But for **instantaneous cooling**, the total heat: \[ Q_{total} = \text{Heat transferred} \approx \text{(conduction through mold wall)} \] Given the problem's data, the main focus is on the **heat flow** during freezing, which is related to the **conduction** through the mold. --- # Step 6: Use thermal conduction to find the mold's thickness The heat conduction rate: \[ Q = \frac{k \times A \times \Delta T}{d} \] Where: - \(k = 1.6 \, W/m\cdot K\) - \(A\) is the surface area of the mold (assuming sphere or cube, but not specified) - \(\Delta T\) is the temperature difference between the mold and liquid metal - \(d\) is the thickness of the mold wall Since the problem states "if the mold-metal surface reaches thermal equilibrium instantly," and the **total heat transfer** is given by the energy balance, we are to find \(d\). --- # Step 7: Final calculation Given the total heat \(Q_{total}\), the heat conduction per unit time: \[ Q_{total} = \frac{k \times A \times (T_{mould} - T_{liquid})}{d} \] Assuming the surface area \(A\) is known or uniform, and the temperature difference is \(\Delta T = T_{mould} - T_{liquid}\). **Since the problem does not specify the mold geometry or surface area,** it appears the intended answer is to find the **mould thickness \(d\)** using the provided data, that satisfies the heat flow: \[ Q_{total} = \frac{k \times A \times \Delta T}{d} \] Rearranged: \[ d = \frac{k \times A \times \Delta T}{Q_{total}} \] **Without explicit \(A\) or \(T_{mould}\),** the typical approach in such problems is to assume the **mould thickness** is determined by the known properties and the heat transfer rate. --- # **Final Answer:** **Given the data, the key is to equate the heat released during cooling and freezing to the conductive heat transfer through the mold wall.** If the surface area \(A\) is normalized or considered unit area: \[ \boxed{ d = \frac{k \times \Delta T}{Q_{total} / A} } \] Since the exact surface area isn't specified, and the problem emphasizes the **instantaneous heat transfer**, the critical conclusion is: \[ \boxed{ \text{Thickenss of the mold } d \propto \frac{k \times \Delta T}{Q_{total}} } \] **In summary:** - The total energy to be removed: approximately **27.477 GJ**. - Heat conduction rate: depends on \(d\), \(k\), and temperature difference. - Final calculation would require the specific surface area to numerically determine \(d\). --- **Note:** The problem as provided appears to be incomplete for a numerical final answer unless additional data (like the mold's surface area or temperature difference) is given. The core concept is understanding the energy balance and conduction relation as outlined.

# Step-by-Step Solution Let's solve the problem step-by-step, focusing on the thermal equilibrium and the properties of the materials involved. --- ## Given Data - **Mould Temperature (\(T_o\))**: 25°C - **Freezing Temperature Range**: 1055°C to 1045°C - **Thermal Conductivity of Mould (\(k_{mould}\))**: 1.6 W/m·K - **Thermal Conductivity of Liquid (\(k_{liquid}\))**: 108 W/m·K - **Density of Mould (\(\rho_{mould}\))**: 3.2 × 1000 kg/m³ - **Density of Liquid (\(\rho_{liquid}\))**: 8.52 × 1000 kg/m³ - **Specific Heat Capacity of Mould (\(C_{p,mould}\))**: 1000 J/kg·K - **Specific Heat Capacity of Liquid (\(C_{p,liquid}\))**: 385 J/kg·K - **Latent Heat of Fusion (\(L_f\))**: 3.2 × 100 kJ/m³ --- ## Part (a): Finding \(T_s\) When Freezing is 0°C ### Step 1: Establishing the Energy Balance When the liquid metal is poured into the mould and reaches thermal equilibrium instantly, we can assume: \[ Q_{lost} = Q_{gained} \] ### Step 2: Heat Lost by Liquid Metal Assuming the liquid metal cools from its pouring temperature (\(T_p\)) to its freezing point (\(T_s\)), we can express the heat lost as: \[ Q_{lost} = m \cdot C_{p,liquid} \cdot (T_p - T_s) + m \cdot L_f \] Here: - \(m\) is the mass of the liquid metal. ### Step 3: Heat Gained by Mould The mould gains heat due to the temperature difference between the liquid metal and itself. The heat gained by the mould is expressed as: \[ Q_{gained} = m_{mould} \cdot C_{p,mould} \cdot (T_s - T_o) \] ### Step 4: Setting Up the Equation At thermal equilibrium: \[ m \cdot C_{p,liquid} \cdot (T_p - T_s) + m \cdot L_f = m_{mould} \cdot C_{p,mould} \cdot (T_s - T_o) \] ### Step 5: Solving for \(T_s\) To find \(T_s\), we need to express \(m\) and \(m_{mould}\) in terms of the densities and volume. Assuming a volume \(V\) of liquid metal: \[ m = \rho_{liquid} \cdot V \] Mould mass could be estimated if its geometry and volume are known, but let's assume \(m_{mould} = \rho_{mould} \cdot V_m\). Plugging in the values: 1. For **Liquid Metal**: \[ Q_{lost} = \rho_{liquid} \cdot V \cdot C_{p,liquid} \cdot (T_p - T_s) + \rho_{liquid} \cdot V \cdot L_f \] 2. For **Mould**: \[ Q_{gained} = \rho_{mould} \cdot V_m \cdot C_{p,mould} \cdot (T_s - T_o) \] Equating \(Q_{lost}\) and \(Q_{gained}\): \[ \rho_{liquid} \cdot V \cdot C_{p,liquid} \cdot (T_p - T_s) + \rho_{liquid} \cdot V \cdot L_f = \rho_{mould} \cdot V_m \cdot C_{p,mould} \cdot (T_s - T_o) \] This equation can be solved for \(T_s\), but we need specific values for \(V\) and \(V_m\). --- ## Part (b): Minimum Pouring Temperature to Prevent Instantaneous Freezing ### Step 1: Freezing Temperature Range The freezing range for the brass alloy is from 1055°C to 1045°C. To prevent instantaneous freezing, the pouring temperature must be above the upper limit of the freezing range. ### Step 2: Minimum Pouring Temperature Calculation To avoid freezing, the pouring temperature \(T_p\) must be: \[ T_p > 1055°C \] ### Final Answer - **(a)** The exact temperature \(T_s\) would depend on solving the derived equations with known mass or volume parameters. - **(b)** The minimum pouring temperature to prevent instantaneous freezing is: \[ \boxed{T_p > 1055°C} \] This ensures that the liquid metal remains in a liquid state upon entering the mould.

# Step-by-Step Solution ## Given Data - **Mould Temperature (\(T_o\))**: 25°C - **Freezing Temperature Range**: 1055°C to 1045°C - **Thermal Conductivity of Mould (\(k_{mould}\))**: 1.6 W/m·K - **Thermal Conductivity of Liquid (\(k_{liquid}\))**: 108 W/m·K - **Density of Mould (\(\rho_{mould}\))**: 3.2 × 1000 kg/m³ - **Density of Liquid (\(\rho_{liquid}\))**: 8.52 × 1000 kg/m³ - **Specific Heat Capacity of Mould (\(C_{p,mould}\))**: 1000 J/kg·K - **Specific Heat Capacity of Liquid (\(C_{p,liquid}\))**: 385 J/kg·K - **Latent Heat of Fusion (\(L_f\))**: 3.2 × 100 kJ/m³ --- ## Part (a): Finding \(T_s\) When Freezing is 0°C ### Step 1: Energy Balance When the liquid metal is poured into the mould and reaches thermal equilibrium: \[ Q_{lost} = Q_{gained} \] ### Step 2: Heat Lost by Liquid Metal The heat lost by the liquid metal can be expressed as: \[ Q_{lost} = m \cdot C_{p,liquid} \cdot (T_p - T_s) + m \cdot L_f \] ### Step 3: Heat Gained by Mould The mould gains heat due to the temperature difference: \[ Q_{gained} = m_{mould} \cdot C_{p,mould} \cdot (T_s - T_o) \] ### Step 4: Setting Up the Equation At thermal equilibrium: \[ m \cdot C_{p,liquid} \cdot (T_p - T_s) + m \cdot L_f = m_{mould} \cdot C_{p,mould} \cdot (T_s - T_o) \] ### Step 5: Expressing Mass in Terms of Volume Assuming the volume \(V\) of the liquid metal: \[ m = \rho_{liquid} \cdot V \] Assuming the mould mass: \[ m_{mould} = \rho_{mould} \cdot V_m \] ### Step 6: Plugging in Values 1. Heat Lost: \[ Q_{lost} = \rho_{liquid} \cdot V \cdot C_{p,liquid} \cdot (T_p - T_s) + \rho_{liquid} \cdot V \cdot L_f \] 2. Heat Gained: \[ Q_{gained} = \rho_{mould} \cdot V_m \cdot C_{p,mould} \cdot (T_s - T_o) \] ### Final Equation Equating \(Q_{lost}\) and \(Q_{gained}\): \[ \rho_{liquid} \cdot V \cdot C_{p,liquid} \cdot (T_p - T_s) + \rho_{liquid} \cdot V \cdot L_f = \rho_{mould} \cdot V_m \cdot C_{p,mould} \cdot (T_s - T_o) \] To solve for \(T_s\), we need values for \(V\) and \(V_m\). --- ## Part (b): Minimum Pouring Temperature to Prevent Instantaneous Freezing ### Step 1: Freezing Temperature Range The freezing range for the brass alloy is from 1055°C to 1045°C. To avoid instantaneous freezing, the pouring temperature must be above the upper limit of the freezing range. ### Step 2: Minimum Pouring Temperature Calculation To prevent freezing: \[ T_p > 1055°C \] ### Final Answers - **(a)** The exact \(T_s\) requires solving the derived equations with known mass or volume parameters. - **(b)** The minimum pouring temperature to prevent instantaneous freezing is: \[ \boxed{T_p > 1055°C} \] This ensures that the liquid metal remains in a liquid state upon entering the mould.

# Step-by-Step Solution ## Given Data - **Mould Temperature (\(T_o\))**: 25°C - **Freezing Temperature Range**: 1055°C to 1045°C - **Thermal Conductivity of Mould (\(k_{mould}\))**: 1.6 W/m·K - **Thermal Conductivity of Liquid (\(k_{liquid}\))**: 108 W/m·K - **Density of Mould (\(\rho_{mould}\))**: 3200 kg/m³ - **Density of Liquid (\(\rho_{liquid}\))**: 8520 kg/m³ - **Specific Heat Capacity of Mould (\(C_{p,mould}\))**: 1000 J/kg·K - **Specific Heat Capacity of Liquid (\(C_{p,liquid}\))**: 385 J/kg·K - **Latent Heat of Fusion (\(L_f\))**: 320 kJ/m³ --- ## Part (a): Finding \(T_s\) When Freezing is 0°C ### Step 1: Understand the Energy Balance When the liquid metal reaches thermal equilibrium with the mould, the heat lost by the liquid metal equals the heat gained by the mould: \[ Q_{lost} = Q_{gained} \] ### Step 2: Heat Lost by Liquid Metal The heat lost by the liquid metal as it cools from its pouring temperature \(T_p\) to the freezing temperature \(T_s\) includes: 1. Cooling the liquid metal to \(T_s\) 2. The latent heat of fusion The equation for heat lost is: \[ Q_{lost} = m \cdot C_{p,liquid} \cdot (T_p - T_s) + m \cdot L_f \] ### Step 3: Heat Gained by Mould The mould gains heat, calculated as: \[ Q_{gained} = m_{mould} \cdot C_{p,mould} \cdot (T_s - T_o) \] ### Step 4: Express Mass in Terms of Volume Assuming the volume \(V\) of the liquid metal: \[ m = \rho_{liquid} \cdot V \] For the mould, if we denote its mass by \(m_{mould}\): \[ m_{mould} = \rho_{mould} \cdot V_{mould} \] ### Step 5: Setting Up the Energy Balance Equation At thermal equilibrium: \[ \rho_{liquid} \cdot V \cdot C_{p,liquid} \cdot (T_p - T_s) + \rho_{liquid} \cdot V \cdot L_f = \rho_{mould} \cdot V_{mould} \cdot C_{p,mould} \cdot (T_s - T_o) \] ### Step 6: Rearranging to Find \(T_s\) To isolate \(T_s\): \[ \rho_{liquid} \cdot V \cdot C_{p,liquid} \cdot (T_p - T_s) + \rho_{liquid} \cdot V \cdot L_f = \rho_{mould} \cdot V_{mould} \cdot C_{p,mould} \cdot (T_s - T_o) \] This becomes: \[ \rho_{liquid} \cdot V \cdot C_{p,liquid} \cdot T_p - \rho_{liquid} \cdot V \cdot C_{p,liquid} \cdot T_s + \rho_{liquid} \cdot V \cdot L_f = \rho_{mould} \cdot V_{mould} \cdot C_{p,mould} \cdot T_s - \rho_{mould} \cdot V_{mould} \cdot C_{p,mould} \cdot T_o \] ### Step 7: Collecting Terms Rearranging gives: \[ T_s \left( \rho_{liquid} \cdot V \cdot C_{p,liquid} + \rho_{mould} \cdot V_{mould} \cdot C_{p,mould} \right) = \rho_{liquid} \cdot V \cdot C_{p,liquid} \cdot T_p + \rho_{mould} \cdot V_{mould} \cdot C_{p,mould} \cdot T_o - \rho_{liquid} \cdot V \cdot L_f \] ### Step 8: Solving for \(T_s\) \[ T_s = \frac{\rho_{liquid} \cdot V \cdot C_{p,liquid} \cdot T_p + \rho_{mould} \cdot V_{mould} \cdot C_{p,mould} \cdot T_o - \rho_{liquid} \cdot V \cdot L_f}{\rho_{liquid} \cdot V \cdot C_{p,liquid} + \rho_{mould} \cdot V_{mould} \cdot C_{p,mould}} \] --- ## Part (b): Minimum Pouring Temperature to Prevent Instantaneous Freezing ### Step 1: Freezing Temperature Range To avoid instantaneous freezing, the pouring temperature \(T_p\) must be higher than the upper limit of the freezing range: \[ T_p > 1055^\circ C \] ### Final Answers - **(a)** The exact temperature \(T_s\) depends on volume parameters, but the general formula is derived. - **(b)** The minimum pouring temperature to prevent instantaneous freezing is: \[ \boxed{T_p > 1055^\circ C} \] This ensures that the liquid metal remains liquid when poured into the mould.