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2) If 0.2 kg of 815° C iron is quenched in a bucket with 5 kg of liquid water at 30° C, what will be the final temperature of the iron and water.

2) If 0.2 kg of 815° C iron is quenched in a bucket with 5 kg of liquid water at 30° C, what will be the final temperature of the iron and water.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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**Problem Statement:** If 0.2 kg of iron at 815°C is quenched in a bucket containing 5 kg of liquid water at 30°C, what will be the final temperature of the iron and water? **Solution Explanation:** To solve this problem, we will apply the principle of conservation of energy. The heat lost by the iron will equal the heat gained by the water as no heat is lost to the surroundings. 1. **Determine the specific heat capacities:** - Specific heat capacity of iron (\(c_{iron}\)) = 0.449 J/g°C - Specific heat capacity of water (\(c_{water}\)) = 4.186 J/g°C 2. **Set up the heat transfer equation:** Let \(T_f\) be the final temperature. \[ m_{iron} \cdot c_{iron} \cdot (T_{initial, iron} - T_f) = m_{water} \cdot c_{water} \cdot (T_f - T_{initial, water}) \] - \(m_{iron}\) = 0.2 kg = 200 g - \(m_{water}\) = 5 kg = 5000 g - \(T_{initial, iron}\) = 815°C - \(T_{initial, water}\) = 30°C 3. **Solve for \(T_f\):** \[ 200 \cdot 0.449 \cdot (815 - T_f) = 5000 \cdot 4.186 \cdot (T_f - 30) \] \[ 89.8 \cdot (815 - T_f) = 20930 \cdot (T_f - 30) \] Simplify and solve the equation to find the final temperature \(T_f\). **Note:** This calculation assumes no heat is lost to the surroundings and complete thermal equilibrium is reached.
Transcribed Image Text:**Problem Statement:** If 0.2 kg of iron at 815°C is quenched in a bucket containing 5 kg of liquid water at 30°C, what will be the final temperature of the iron and water? **Solution Explanation:** To solve this problem, we will apply the principle of conservation of energy. The heat lost by the iron will equal the heat gained by the water as no heat is lost to the surroundings. 1. **Determine the specific heat capacities:** - Specific heat capacity of iron (\(c_{iron}\)) = 0.449 J/g°C - Specific heat capacity of water (\(c_{water}\)) = 4.186 J/g°C 2. **Set up the heat transfer equation:** Let \(T_f\) be the final temperature. \[ m_{iron} \cdot c_{iron} \cdot (T_{initial, iron} - T_f) = m_{water} \cdot c_{water} \cdot (T_f - T_{initial, water}) \] - \(m_{iron}\) = 0.2 kg = 200 g - \(m_{water}\) = 5 kg = 5000 g - \(T_{initial, iron}\) = 815°C - \(T_{initial, water}\) = 30°C 3. **Solve for \(T_f\):** \[ 200 \cdot 0.449 \cdot (815 - T_f) = 5000 \cdot 4.186 \cdot (T_f - 30) \] \[ 89.8 \cdot (815 - T_f) = 20930 \cdot (T_f - 30) \] Simplify and solve the equation to find the final temperature \(T_f\). **Note:** This calculation assumes no heat is lost to the surroundings and complete thermal equilibrium is reached.
Expert Solution
Step 1

According to the conservation of energy, 

"The total energy of an isolated system always remains constant"

Therefore, the heat loose by the iron will be equal to the heat gain by the water.

Use the data table:

The specific heat of water (Cw) = 4.182 kJ/kg.K

The specific heat of iron (Ci) = 0.45 kJ/kg.K

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