(a) In reaching equilibrium, how much heat transfer occurs from 1.2 kg of water at 40°C when it is placed in contact with 1.2 kg of 20°C water? Specific heat of water c=4186 J/(kg°C) Hint: If the masses of water are equal, what is the equilirium temperature of the water mixture? Q = 50232 (b) What is the change in entropy due to this heat transfer? AS total = J/K (c) How much work is made unavailable, taking the lowest temperature to be 20°C?
(a) In reaching equilibrium, how much heat transfer occurs from 1.2 kg of water at 40°C when it is placed in contact with 1.2 kg of 20°C water? Specific heat of water c=4186 J/(kg°C) Hint: If the masses of water are equal, what is the equilirium temperature of the water mixture? Q = 50232 (b) What is the change in entropy due to this heat transfer? AS total = J/K (c) How much work is made unavailable, taking the lowest temperature to be 20°C?
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Qs b and c
![### Thermodynamics Problem: Heat Transfer and Entropy Change
#### Problem Statement
**(a)** In reaching equilibrium, how much heat transfer occurs from 1.2 kg of water at 40°C when it is placed in contact with 1.2 kg of 20°C water? The specific heat of water is given as \(c = 4186 \, \text{J/(kg°C)}\).
*Hint*: If the masses of water are equal, what is the equilibrium temperature of the water mixture?
\[ Q = \boxed{50232 \, \text{J}} \]
**(b)** What is the change in entropy due to this heat transfer?
\[ \Delta S_{\text{total}} = \boxed{} \, \text{J/K} \]
**(c)** How much work is made unavailable, taking the lowest temperature to be 20°C?
\[ W_{\text{unavailable}} = \boxed{} \, \text{J} \]
#### Steps to Solve:
**(a) Calculation of Heat Transfer:**
To find the heat transfer \(Q\), we use the equation:
\[ Q = m \cdot c \cdot \Delta T \]
where:
- \(m\) is the mass of water,
- \(c\) is the specific heat of water, and
- \(\Delta T\) is the change in temperature.
Since both masses are equal and the system reaches thermal equilibrium, the equilibrium temperature \(T_{\text{eq}}\) would simply be the average of the initial temperatures:
\[ T_{\text{eq}} = \frac{40^\circ\text{C} + 20^\circ \text{C}}{2} = 30^\circ \text{C} \]
Thus, the change in temperature for each mass is:
\[\Delta T = |T_{\text{initial}} - T_{\text{eq}}| = |40^\circ\text{C} - 30^\circ \text{C}| = 10^\circ \text{C}\]
Given specific values:
\[m = 1.2 \, \text{kg}, \quad c = 4186 \, \text{J/(kg°C)}\]
So:
\[ Q = 1.2 \, \text{kg} \times 4186 \, \text{J/(kg°C)} \times 10](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff787f97c-9dc4-4f39-97ba-c4c8bbc1d69b%2F86309929-eb4e-4c75-a756-467b421cae5b%2Fsq0uj5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Thermodynamics Problem: Heat Transfer and Entropy Change
#### Problem Statement
**(a)** In reaching equilibrium, how much heat transfer occurs from 1.2 kg of water at 40°C when it is placed in contact with 1.2 kg of 20°C water? The specific heat of water is given as \(c = 4186 \, \text{J/(kg°C)}\).
*Hint*: If the masses of water are equal, what is the equilibrium temperature of the water mixture?
\[ Q = \boxed{50232 \, \text{J}} \]
**(b)** What is the change in entropy due to this heat transfer?
\[ \Delta S_{\text{total}} = \boxed{} \, \text{J/K} \]
**(c)** How much work is made unavailable, taking the lowest temperature to be 20°C?
\[ W_{\text{unavailable}} = \boxed{} \, \text{J} \]
#### Steps to Solve:
**(a) Calculation of Heat Transfer:**
To find the heat transfer \(Q\), we use the equation:
\[ Q = m \cdot c \cdot \Delta T \]
where:
- \(m\) is the mass of water,
- \(c\) is the specific heat of water, and
- \(\Delta T\) is the change in temperature.
Since both masses are equal and the system reaches thermal equilibrium, the equilibrium temperature \(T_{\text{eq}}\) would simply be the average of the initial temperatures:
\[ T_{\text{eq}} = \frac{40^\circ\text{C} + 20^\circ \text{C}}{2} = 30^\circ \text{C} \]
Thus, the change in temperature for each mass is:
\[\Delta T = |T_{\text{initial}} - T_{\text{eq}}| = |40^\circ\text{C} - 30^\circ \text{C}| = 10^\circ \text{C}\]
Given specific values:
\[m = 1.2 \, \text{kg}, \quad c = 4186 \, \text{J/(kg°C)}\]
So:
\[ Q = 1.2 \, \text{kg} \times 4186 \, \text{J/(kg°C)} \times 10
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