A coffee cup calorimeter contains 156.02 g of water at 22.50 °C. A 65.468 g piece of iron is heated to 100.44 °C. The piece of iron is added to the coffee cup caloriemter and the contents reach thermal equilibrium at 25.68 °C. The specific heat capacity of iron is 0.449 J and the g-K specific heat capacity of water is 4.184 g.K How much heat, q, is lost by the piece of iron? Giron = How much heat, q, is gained by the water? 9water =
A coffee cup calorimeter contains 156.02 g of water at 22.50 °C. A 65.468 g piece of iron is heated to 100.44 °C. The piece of iron is added to the coffee cup caloriemter and the contents reach thermal equilibrium at 25.68 °C. The specific heat capacity of iron is 0.449 J and the g-K specific heat capacity of water is 4.184 g.K How much heat, q, is lost by the piece of iron? Giron = How much heat, q, is gained by the water? 9water =
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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Coffee cup caloriemeter homework..thanks
![A coffee cup calorimeter contains 156.02 g of water at 22.50 °C. A 65.468 g piece of iron is heated to 100.44 °C. The piece of iron is added to the coffee cup calorimeter and the contents reach thermal equilibrium at 25.68 °C. The specific heat capacity of iron is 0.449 J/g·K and the specific heat capacity of water is 4.184 J/g·K.
How much heat, \( q \), is lost by the piece of iron?
\( q_{\text{iron}} = \)
How much heat, \( q \), is gained by the water?
\( q_{\text{water}} = \)
---
Explanation:
- The specific heat capacity is a property that tells us how much heat energy is needed to change the temperature of a substance by 1 K (or 1 °C).
- The process described involves thermal equilibrium, where the heat lost by the iron will equal the heat gained by the water assuming no heat loss to the surroundings.
- Calculating the heat transferred involves using the formula:
\[
q = m \times c \times \Delta T
\]
where \( q \) is the heat transfer, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6ae9165-8c64-40e9-9a68-65bc31ca98b7%2F82e16119-6a44-4547-9e9c-67aeefaf52c4%2F3dmrhoni_processed.png&w=3840&q=75)
Transcribed Image Text:A coffee cup calorimeter contains 156.02 g of water at 22.50 °C. A 65.468 g piece of iron is heated to 100.44 °C. The piece of iron is added to the coffee cup calorimeter and the contents reach thermal equilibrium at 25.68 °C. The specific heat capacity of iron is 0.449 J/g·K and the specific heat capacity of water is 4.184 J/g·K.
How much heat, \( q \), is lost by the piece of iron?
\( q_{\text{iron}} = \)
How much heat, \( q \), is gained by the water?
\( q_{\text{water}} = \)
---
Explanation:
- The specific heat capacity is a property that tells us how much heat energy is needed to change the temperature of a substance by 1 K (or 1 °C).
- The process described involves thermal equilibrium, where the heat lost by the iron will equal the heat gained by the water assuming no heat loss to the surroundings.
- Calculating the heat transferred involves using the formula:
\[
q = m \times c \times \Delta T
\]
where \( q \) is the heat transfer, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.
![**Heat Transfer in a Calorimeter Experiment**
In this experiment, we investigate the heat transfer between a piece of iron and water, considering the heat absorbed by a styrofoam calorimeter.
**Objective:**
1. Determine the heat capacity of the styrofoam calorimeter in joules per kelvin (J/K).
2. Calculate the final temperature of the system, assuming all the heat lost by the iron is absorbed by the water.
**Problems:**
1. What is the heat capacity of the styrofoam calorimeter in joules per kelvin \(\left(\frac{J}{K}\right)\)?
\[ C_{\text{calorimeter}} = \]
2. What would be the final temperature of the system if all the heat lost by the iron was absorbed by the water?
\[ T_{\text{final}} = \]
**Instructions:**
- Consider the heat lost by the iron, the heat gained by the water, and the potential heat absorbed by the calorimeter.
- Use relevant equations for heat transfer and calorimetry to solve the problems.
These tasks enhance understanding of heat transfer principles and improve problem-solving skills in thermodynamics.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6ae9165-8c64-40e9-9a68-65bc31ca98b7%2F82e16119-6a44-4547-9e9c-67aeefaf52c4%2Fjs85smp_processed.png&w=3840&q=75)
Transcribed Image Text:**Heat Transfer in a Calorimeter Experiment**
In this experiment, we investigate the heat transfer between a piece of iron and water, considering the heat absorbed by a styrofoam calorimeter.
**Objective:**
1. Determine the heat capacity of the styrofoam calorimeter in joules per kelvin (J/K).
2. Calculate the final temperature of the system, assuming all the heat lost by the iron is absorbed by the water.
**Problems:**
1. What is the heat capacity of the styrofoam calorimeter in joules per kelvin \(\left(\frac{J}{K}\right)\)?
\[ C_{\text{calorimeter}} = \]
2. What would be the final temperature of the system if all the heat lost by the iron was absorbed by the water?
\[ T_{\text{final}} = \]
**Instructions:**
- Consider the heat lost by the iron, the heat gained by the water, and the potential heat absorbed by the calorimeter.
- Use relevant equations for heat transfer and calorimetry to solve the problems.
These tasks enhance understanding of heat transfer principles and improve problem-solving skills in thermodynamics.
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