4) Calculate the final temperature of water if you mix one liter of water at 40 °C with two liters of water at 20 °C. Density of water is 1000 kg/m³.
4) Calculate the final temperature of water if you mix one liter of water at 40 °C with two liters of water at 20 °C. Density of water is 1000 kg/m³.
Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
8th Edition
ISBN:9781305387102
Author:Kreith, Frank; Manglik, Raj M.
Publisher:Kreith, Frank; Manglik, Raj M.
Chapter6: Forced Convection Over Exterior Surfaces
Section: Chapter Questions
Problem 6.25P
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4.
![**4) Calculate the final temperature of water if you mix one liter of water at 40 °C with two liters of water at 20 °C. Density of water is 1000 kg/m³.**
**Solution:**
To find the final temperature (\(T_f\)) when two volumes of water at different temperatures are mixed, we can use the concept of conservation of energy. The heat lost by the warmer water will equal the heat gained by the cooler water.
Let:
- \(m_1 = 1 \text{ kg}\) (mass of 1 liter of water at 40 °C)
- \(T_1 = 40 °C\) (initial temperature of the first sample)
- \(m_2 = 2 \text{ kg}\) (mass of 2 liters of water at 20 °C)
- \(T_2 = 20 °C\) (initial temperature of the second sample)
Since the density of water is 1000 kg/m³, 1 liter of water has a mass of 1 kg.
**Formula:**
\[ m_1 \cdot c \cdot (T_f - T_1) + m_2 \cdot c \cdot (T_f - T_2) = 0 \]
where \(c\) is the specific heat capacity of water, approximately 4.18 kJ/(kg·°C). Since \(c\) is constant for both water samples, it cancels out, simplifying to:
\[ m_1 \cdot (T_f - T_1) + m_2 \cdot (T_f - T_2) = 0 \]
**Equation:**
\[ 1 \cdot (T_f - 40) + 2 \cdot (T_f - 20) = 0 \]
**Solve for \(T_f\):**
Expanding and combining terms:
\[ T_f - 40 + 2T_f - 40 = 0 \]
\[ 3T_f - 80 = 0 \]
\[ 3T_f = 80 \]
\[ T_f = \frac{80}{3} \]
\[ T_f \approx 26.67 \, °C \]
Thus, the final temperature of the mixture is approximately 26.67 °C.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd7323f76-5a9c-46b6-9930-86cefe6607d1%2Ff961d5ac-9aa6-40bf-9b08-98a4d9c58916%2F111g54_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**4) Calculate the final temperature of water if you mix one liter of water at 40 °C with two liters of water at 20 °C. Density of water is 1000 kg/m³.**
**Solution:**
To find the final temperature (\(T_f\)) when two volumes of water at different temperatures are mixed, we can use the concept of conservation of energy. The heat lost by the warmer water will equal the heat gained by the cooler water.
Let:
- \(m_1 = 1 \text{ kg}\) (mass of 1 liter of water at 40 °C)
- \(T_1 = 40 °C\) (initial temperature of the first sample)
- \(m_2 = 2 \text{ kg}\) (mass of 2 liters of water at 20 °C)
- \(T_2 = 20 °C\) (initial temperature of the second sample)
Since the density of water is 1000 kg/m³, 1 liter of water has a mass of 1 kg.
**Formula:**
\[ m_1 \cdot c \cdot (T_f - T_1) + m_2 \cdot c \cdot (T_f - T_2) = 0 \]
where \(c\) is the specific heat capacity of water, approximately 4.18 kJ/(kg·°C). Since \(c\) is constant for both water samples, it cancels out, simplifying to:
\[ m_1 \cdot (T_f - T_1) + m_2 \cdot (T_f - T_2) = 0 \]
**Equation:**
\[ 1 \cdot (T_f - 40) + 2 \cdot (T_f - 20) = 0 \]
**Solve for \(T_f\):**
Expanding and combining terms:
\[ T_f - 40 + 2T_f - 40 = 0 \]
\[ 3T_f - 80 = 0 \]
\[ 3T_f = 80 \]
\[ T_f = \frac{80}{3} \]
\[ T_f \approx 26.67 \, °C \]
Thus, the final temperature of the mixture is approximately 26.67 °C.
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