27) A person tries to heat up her bath water by adding 5.0 L of water at 80°C to 60 L of water at 30°C. What is the final temperature of the bath water? A) 40°C B) 36°C C) 34°C D) 38°C
27) A person tries to heat up her bath water by adding 5.0 L of water at 80°C to 60 L of water at 30°C. What is the final temperature of the bath water? A) 40°C B) 36°C C) 34°C D) 38°C
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Chapter1: Units, Trigonometry. And Vectors
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Please show all calculations and type them out since I have bad eyesight and it is very difficult for me to read handwriting
![### Problem Statement
A person tries to heat up her bath water by adding 5.0 L of water at 80°C to 60 L of water at 30°C. What is the final temperature of the bath water?
#### Possible Answers:
A) 40°C
B) 36°C
C) 34°C
D) 38°C
This question involves understanding the concept of heat transfer and equilibrium temperature. To solve the problem, we need to apply the principle of conservation of energy.
Here’s a brief overview of how to approach such a problem:
1. **Identify the volumes and temperatures**:
- Volume of hot water, V₁ = 5.0 L at T₁ = 80°C
- Volume of cold water, V₂ = 60.0 L at T₂ = 30°C
2. **Assumption**: Since it deals with water and assuming no heat loss to the environment, we can use the formula for mixing:
\[ T_f = \frac{(V_1 \cdot T_1) + (V_2 \cdot T_2)}{V_1 + V_2} \]
Where,
- \( T_f \) = final temperature
- \( V_1 \) and \( V_2 \) are the volumes of water
- \( T_1 \) and \( T_2 \) are the initial temperatures of the water volumes.
3. **Calculate**:
Plugging in the values:
\[ T_f = \frac{(5.0 \cdot 80) + (60.0 \cdot 30)}{5.0 + 60.0} = \frac{(400) + (1800)}{65} = \frac{2200}{65} \approx 34°C \]
Thus, the final temperature of the bath water is **34°C**.
#### Correct Answer:
C) 34°C](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F93bd9d77-dcfb-4bce-a8b6-6527873699f2%2Fdb022f9b-5e98-4dd1-82a9-ae124f3504f1%2Fpdaq51_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A person tries to heat up her bath water by adding 5.0 L of water at 80°C to 60 L of water at 30°C. What is the final temperature of the bath water?
#### Possible Answers:
A) 40°C
B) 36°C
C) 34°C
D) 38°C
This question involves understanding the concept of heat transfer and equilibrium temperature. To solve the problem, we need to apply the principle of conservation of energy.
Here’s a brief overview of how to approach such a problem:
1. **Identify the volumes and temperatures**:
- Volume of hot water, V₁ = 5.0 L at T₁ = 80°C
- Volume of cold water, V₂ = 60.0 L at T₂ = 30°C
2. **Assumption**: Since it deals with water and assuming no heat loss to the environment, we can use the formula for mixing:
\[ T_f = \frac{(V_1 \cdot T_1) + (V_2 \cdot T_2)}{V_1 + V_2} \]
Where,
- \( T_f \) = final temperature
- \( V_1 \) and \( V_2 \) are the volumes of water
- \( T_1 \) and \( T_2 \) are the initial temperatures of the water volumes.
3. **Calculate**:
Plugging in the values:
\[ T_f = \frac{(5.0 \cdot 80) + (60.0 \cdot 30)}{5.0 + 60.0} = \frac{(400) + (1800)}{65} = \frac{2200}{65} \approx 34°C \]
Thus, the final temperature of the bath water is **34°C**.
#### Correct Answer:
C) 34°C
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