A 496−g piece of copper tubing is heated to 89.5 ° C and placed in an insulated vessel containing 159 g of water at 22.8 ° C. Assuming no loss of water and a heat capacity for the vessel of 10.0 J/ ° C, what is the final temperature of the system (c of copper = 0.387 J/g · ° C)?
A 496−g piece of copper tubing is heated to 89.5 ° C and placed in an insulated vessel containing 159 g of water at 22.8 ° C. Assuming no loss of water and a heat capacity for the vessel of 10.0 J/ ° C, what is the final temperature of the system (c of copper = 0.387 J/g · ° C)?
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
A 496−g piece of copper tubing is heated to 89.5
°
C and placed in an insulated vessel containing
159 g
of water at 22.8
°
C. Assuming no loss of water and a heat capacity for the vessel of 10.0 J/
°
C, what is the final temperature of the system (c of copper = 0.387 J/g
·
°
C)?
![**Problem Statement:**
What is the change in internal energy (in J) of a system that absorbs 0.413 kJ of heat from its surroundings and has 0.779 kcal of work done on it? Give your answer in scientific notation.
**Answer Format:**
\[ \boxed{} \times 10^{\boxed{}} \, \text{J} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F04982bc8-2af1-43a7-9ca8-af3a0a4dba7d%2F8a3ae30c-fc98-4855-8a6a-fdc3c342c02a%2Fi4r37_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
What is the change in internal energy (in J) of a system that absorbs 0.413 kJ of heat from its surroundings and has 0.779 kcal of work done on it? Give your answer in scientific notation.
**Answer Format:**
\[ \boxed{} \times 10^{\boxed{}} \, \text{J} \]
![**Problem:**
A 376-g aluminum engine part at an initial temperature of 19.29°C absorbs 57.7 kJ of heat. What is the final temperature of the part (c of Al = 0.900 J/g·K)?
**Solution:**
To find the final temperature of the aluminum engine part, we use the formula:
\[ q = mc\Delta T \]
where:
- \( q \) is the heat absorbed (57.7 kJ or 57700 J),
- \( m \) is the mass (376 g),
- \( c \) is the specific heat capacity (0.900 J/g·K),
- \( \Delta T \) is the change in temperature (\( T_{final} - T_{initial} \)).
Rearranging the formula to solve for the final temperature:
\[ \Delta T = \frac{q}{mc} \]
\[ T_{final} = T_{initial} + \Delta T \]
Substitute the values:
\[ \Delta T = \frac{57700}{376 \times 0.900} \]
\[ T_{final} = 19.29 + \Delta T \]
Calculate \( \Delta T \) and then \( T_{final} \).
**Answer:**
\[ \Delta T = \frac{57700}{338.4} \approx 170.56 \]
\[ T_{final} = 19.29 + 170.56 \approx 189.85°C \]
**Diagram Explanation:**
In this problem, there is no diagram or graph to explain. The problem deals with mathematical calculations based on provided physical values.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F04982bc8-2af1-43a7-9ca8-af3a0a4dba7d%2F8a3ae30c-fc98-4855-8a6a-fdc3c342c02a%2Fxe0vt8o_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:**
A 376-g aluminum engine part at an initial temperature of 19.29°C absorbs 57.7 kJ of heat. What is the final temperature of the part (c of Al = 0.900 J/g·K)?
**Solution:**
To find the final temperature of the aluminum engine part, we use the formula:
\[ q = mc\Delta T \]
where:
- \( q \) is the heat absorbed (57.7 kJ or 57700 J),
- \( m \) is the mass (376 g),
- \( c \) is the specific heat capacity (0.900 J/g·K),
- \( \Delta T \) is the change in temperature (\( T_{final} - T_{initial} \)).
Rearranging the formula to solve for the final temperature:
\[ \Delta T = \frac{q}{mc} \]
\[ T_{final} = T_{initial} + \Delta T \]
Substitute the values:
\[ \Delta T = \frac{57700}{376 \times 0.900} \]
\[ T_{final} = 19.29 + \Delta T \]
Calculate \( \Delta T \) and then \( T_{final} \).
**Answer:**
\[ \Delta T = \frac{57700}{338.4} \approx 170.56 \]
\[ T_{final} = 19.29 + 170.56 \approx 189.85°C \]
**Diagram Explanation:**
In this problem, there is no diagram or graph to explain. The problem deals with mathematical calculations based on provided physical values.
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