Chambers Beck
05/07/2023 · Elementary School
How many joules of heat are required to heat 148.3 grams of room temperature water \( \left(22^{\circ} \mathrm{C}\right) \) to the boiling point? Answer: Check
UpStudy ThothAI Solution
Tutor-Verified Answer
Step-by-step Solution
To calculate the amount of heat required to heat 148.3 grams of water from room temperature (\(22^{\circ} \mathrm{C}\)) to the boiling point (\(100^{\circ} \mathrm{C}\)), we can use the formula for heat transfer:
\[ Q = m \cdot c \cdot \Delta T \]
where:
- \( Q \) is the heat energy in joules (J),
- \( m \) is the mass of the water in grams (g),
- \( c \) is the specific heat capacity of water in \( \frac{J}{g \cdot ^{\circ}C} \),
- \( \Delta T \) is the change in temperature in degrees Celsius (\(^{\circ}C\)).
Given:
- Mass of water, \( m = 148.3 \, \text{g} \),
- Specific heat capacity of water, \( c = 4.18 \, \frac{J}{g \cdot ^{\circ}C} \),
- Initial temperature, \( T_{\text{initial}} = 22^{\circ} \mathrm{C} \),
- Final temperature, \( T_{\text{final}} = 100^{\circ} \mathrm{C} \).
First, we need to calculate the change in temperature (\( \Delta T \)):
\[ \Delta T = T_{\text{final}} - T_{\text{initial}} \]
Then, we can substitute the values into the formula to find the heat energy (\( Q \)) in joules.
Calculate the value by following steps:
- step0: Calculate:
\(100-22\)
- step1: Subtract the numbers:
\(78\)
The change in temperature (\( \Delta T \)) is 78 degrees Celsius.
Now, we can calculate the heat energy (\( Q \)) required to heat 148.3 grams of water from \(22^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) using the formula:
\[ Q = m \cdot c \cdot \Delta T \]
Substitute the values:
- Mass of water, \( m = 148.3 \, \text{g} \),
- Specific heat capacity of water, \( c = 4.18 \, \frac{J}{g \cdot ^{\circ}C} \),
- Change in temperature, \( \Delta T = 78 \, ^{\circ}C \).
Let's calculate the heat energy (\( Q \)) in joules.
Calculate the value by following steps:
- step0: Calculate:
\(148.3\times 4.18\times 78\)
- step1: Multiply the terms:
\(619.894\times 78\)
- step2: Multiply the numbers:
\(48351.732\)
The amount of heat required to heat 148.3 grams of water from \(22^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) is approximately 48351.732 joules.
Quick Answer
The amount of heat required is approximately 48351.732 joules.
Answered by UpStudy AI and reviewed by a Professional Tutor
UpStudy ThothAI
Self-Developed and Ever-Improving
Thoth AI product is constantly being upgraded and optimized.
Covers All Major Subjects
Capable of handling homework in math, chemistry, biology, physics, and more.
Instant and Accurate
Provides immediate and precise solutions and guidance.
Try Now
Ask Tutors
Ask AI
10x
Fastest way to Get Answers & Solutions
By text
Enter your question here…
By image
Re-Upload
Submit