25) In a flask, 114.0 g of water is heated using 67.0 W of power, with perfect efficiency. How long will it take to raise the temperature of the water from 15°C to 25°C? The specific heat of water is 4186J/kg.K. A) 17 s B) 71 s C) 320,000 s D) 4.1 s
25) In a flask, 114.0 g of water is heated using 67.0 W of power, with perfect efficiency. How long will it take to raise the temperature of the water from 15°C to 25°C? The specific heat of water is 4186J/kg.K. A) 17 s B) 71 s C) 320,000 s D) 4.1 s
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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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![### Physics Problem: Heating Water in a Flask
**Problem Statement:**
In a flask, 114.0 g of water is heated using 67.0 W of power, with perfect efficiency. How long will it take to raise the temperature of the water from 15°C to 25°C? The specific heat of water is 4186 J/kg·K.
**Options:**
- A) 17 s
- B) 71 s
- C) 320,000 s
- D) 4.1 s
---
This problem requires the application of the formula for heat transfer, \( Q = mc\Delta T \), where:
- \( Q \) is the heat energy,
- \( m \) is the mass of the water,
- \( c \) is the specific heat capacity,
- \( \Delta T \) is the change in temperature.
We also use the relation involving power, \( P = \frac{Q}{t} \), where \( P \) is power and \( t \) is time, which needs to be solved.
### Steps to Solve:
1. **Convert mass from grams to kilograms:**
\[
m = 114.0 \, \text{g} = 0.114 \, \text{kg}
\]
2. **Calculate the temperature change:**
\[
\Delta T = 25^\circ \text{C} - 15^\circ \text{C} = 10^\circ \text{C}
\]
3. **Calculate the heat energy required:**
\[
Q = mc\Delta T = 0.114 \, \text{kg} \times 4186 \, \text{J/kg·K} \times 10 \, \text{K}
\]
\[
Q = 4762.44 \, \text{J}
\]
4. **Solve for time using the power equation:**
\[
P = \frac{Q}{t} \implies t = \frac{Q}{P}
\]
\[
t = \frac{4762.44 \, \text{J}}{67.0 \, \text{W}}
\]
\[
t \approx 71.1 \, \text{s}
\]
Therefore, the correct](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F93bd9d77-dcfb-4bce-a8b6-6527873699f2%2Fdda94459-79f8-4205-8c52-c50a087aab8c%2F61uspyf_processed.png&w=3840&q=75)
Transcribed Image Text:### Physics Problem: Heating Water in a Flask
**Problem Statement:**
In a flask, 114.0 g of water is heated using 67.0 W of power, with perfect efficiency. How long will it take to raise the temperature of the water from 15°C to 25°C? The specific heat of water is 4186 J/kg·K.
**Options:**
- A) 17 s
- B) 71 s
- C) 320,000 s
- D) 4.1 s
---
This problem requires the application of the formula for heat transfer, \( Q = mc\Delta T \), where:
- \( Q \) is the heat energy,
- \( m \) is the mass of the water,
- \( c \) is the specific heat capacity,
- \( \Delta T \) is the change in temperature.
We also use the relation involving power, \( P = \frac{Q}{t} \), where \( P \) is power and \( t \) is time, which needs to be solved.
### Steps to Solve:
1. **Convert mass from grams to kilograms:**
\[
m = 114.0 \, \text{g} = 0.114 \, \text{kg}
\]
2. **Calculate the temperature change:**
\[
\Delta T = 25^\circ \text{C} - 15^\circ \text{C} = 10^\circ \text{C}
\]
3. **Calculate the heat energy required:**
\[
Q = mc\Delta T = 0.114 \, \text{kg} \times 4186 \, \text{J/kg·K} \times 10 \, \text{K}
\]
\[
Q = 4762.44 \, \text{J}
\]
4. **Solve for time using the power equation:**
\[
P = \frac{Q}{t} \implies t = \frac{Q}{P}
\]
\[
t = \frac{4762.44 \, \text{J}}{67.0 \, \text{W}}
\]
\[
t \approx 71.1 \, \text{s}
\]
Therefore, the correct
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