Sure! Let's solve this step-by-step:
1. Understanding the Problem:
- We have two portions of water to mix:
- The first portion has 10 kg of water at 25°C.
- The second portion has 4 kg of water at 80°C.
- We need to find the final temperature of the mixture, assuming no heat is lost to the surroundings.
2. Principle to Use:
- We will use the principle of conservation of energy, which states that the total heat lost by the hotter portion of water will be equal to the total heat gained by the cooler portion of water.
3. Formulate the Equation:
- Let [tex]\( T_f \)[/tex] be the final temperature of the mixture.
- The heat lost by the hot water (4 kg at 80°C) can be expressed as: [tex]\( Q_{hot} = m_2 \cdot (T_2 - T_f) \)[/tex]
- The heat gained by the cold water (10 kg at 25°C) can be expressed as: [tex]\( Q_{cold} = m_1 \cdot (T_f - T_1) \)[/tex]
- According to the conservation of energy:
[tex]\[
m_2 \cdot (T_2 - T_f) = m_1 \cdot (T_f - T_1)
\][/tex]
4. Substitute the Values:
- Where:
- [tex]\( m_1 = 10 \, \text{kg} \)[/tex]
- [tex]\( T_1 = 25 \, \text{°C} \)[/tex]
- [tex]\( m_2 = 4 \, \text{kg} \)[/tex]
- [tex]\( T_2 = 80 \, \text{°C} \)[/tex]
- Substitute these values into the equation:
[tex]\[
4 \cdot (80 - T_f) = 10 \cdot (T_f - 25)
\][/tex]
5. Solve for [tex]\( T_f \)[/tex]:
- Expand and simplify the equation:
[tex]\[
320 - 4T_f = 10T_f - 250
\][/tex]
- Combine like terms:
[tex]\[
320 + 250 = 10T_f + 4T_f
\][/tex]
[tex]\[
570 = 14T_f
\][/tex]
- Solve for [tex]\( T_f \)[/tex]:
[tex]\[
T_f = \frac{570}{14}
\][/tex]
[tex]\[
T_f \approx 40.714285714285715 \, \text{°C}
\][/tex]
6. Conclusion:
- Therefore, the final temperature of the mixture when 4 kg of water at 80°C is mixed with 10 kg of water at 25°C, neglecting heat loss to the surroundings, is approximately 40.71 °C.
