13) A hot cup of tea has a mass of 50g with a temperature of 90°C. A person decides to pour 5g of cold water at 10°C. Determine the equilibrium temperature. (Cwater = 1 cal/g°C) Hint: Qhot = -Qcold Show work. A) 82.7°C B) 90°C C) 50°C D) 25°C E) None of the above.
13) A hot cup of tea has a mass of 50g with a temperature of 90°C. A person decides to pour 5g of cold water at 10°C. Determine the equilibrium temperature. (Cwater = 1 cal/g°C) Hint: Qhot = -Qcold Show work. A) 82.7°C B) 90°C C) 50°C D) 25°C E) None of the above.
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Please answer 13 and 14
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### Physics Problems and Solutions
#### Problem 13
A hot cup of tea has a mass of 50g with a temperature of 90°C. A person decides to pour 5g of cold water at 10°C. Determine the equilibrium temperature.
\[ c_{\text{water}} = 1 \text{ cal/g}°C \]
**Hint**: \( Q_{\text{hot}} = - Q_{\text{cold}} \)
**Show work**.
- A) 82.7°C
- B) 90°C
- C) 50°C
- D) 25°C
- E) None of the above.
#### Problem 14
A floating leaf oscillates up and down four complete cycles in one second as a water wave passes by. The wave's wavelength is 7.5 meters. What is the wave's speed?
**Show work**.
- A) 7.5 m/s
- B) 15 m/s
- C) 4 m/s
- D) 30 m/s
#### Explanation
**For Problem 13**
To find the equilibrium temperature (\(T_{\text{final}}\)) we use the principle of conservation of energy. The heat lost by the hot tea will be equal to the heat gained by the cold water:
\[ Q_{\text{hot}} = - Q_{\text{cold}} \]
\[ m_{\text{hot}} \times c_{\text{water}} \times (T_{\text{initial hot}} - T_{\text{final}}) = m_{\text{cold}} \times c_{\text{water}} \times (T_{\text{final}} - T_{\text{initial cold}}) \]
Let:
- \( m_{\text{hot}} = 50g \)
- \( m_{\text{cold}} = 5g \)
- \( c_{\text{water}} = 1 \text{cal/g}°C \)
- \( T_{\text{initial hot}} = 90°C \)
- \( T_{\text{initial cold}} = 10°C \)
Substituting the values, we get:
\[ 50 \times (90 - T_{\text{final}}) = 5 \times (T_{\text{final}} - 10) \]
Solving for \( T_{\text{final}} \):
\[ 4500 - 50 T](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbc6b83f6-6fd0-4bf2-b4c8-4ee11df5765c%2Fcae11156-283b-48e9-ad9b-1cf26bc3666c%2F9tstoa_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
### Physics Problems and Solutions
#### Problem 13
A hot cup of tea has a mass of 50g with a temperature of 90°C. A person decides to pour 5g of cold water at 10°C. Determine the equilibrium temperature.
\[ c_{\text{water}} = 1 \text{ cal/g}°C \]
**Hint**: \( Q_{\text{hot}} = - Q_{\text{cold}} \)
**Show work**.
- A) 82.7°C
- B) 90°C
- C) 50°C
- D) 25°C
- E) None of the above.
#### Problem 14
A floating leaf oscillates up and down four complete cycles in one second as a water wave passes by. The wave's wavelength is 7.5 meters. What is the wave's speed?
**Show work**.
- A) 7.5 m/s
- B) 15 m/s
- C) 4 m/s
- D) 30 m/s
#### Explanation
**For Problem 13**
To find the equilibrium temperature (\(T_{\text{final}}\)) we use the principle of conservation of energy. The heat lost by the hot tea will be equal to the heat gained by the cold water:
\[ Q_{\text{hot}} = - Q_{\text{cold}} \]
\[ m_{\text{hot}} \times c_{\text{water}} \times (T_{\text{initial hot}} - T_{\text{final}}) = m_{\text{cold}} \times c_{\text{water}} \times (T_{\text{final}} - T_{\text{initial cold}}) \]
Let:
- \( m_{\text{hot}} = 50g \)
- \( m_{\text{cold}} = 5g \)
- \( c_{\text{water}} = 1 \text{cal/g}°C \)
- \( T_{\text{initial hot}} = 90°C \)
- \( T_{\text{initial cold}} = 10°C \)
Substituting the values, we get:
\[ 50 \times (90 - T_{\text{final}}) = 5 \times (T_{\text{final}} - 10) \]
Solving for \( T_{\text{final}} \):
\[ 4500 - 50 T
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