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6. For a calorimeter, if a ball with a mass of 0.2 kg, has an initial temperature of 38 °C when placed in a calorimeter filled with water, what is the final temperature (in degrees. Celsius) of the ball if the ball absorbs 33 J from the water. Specific heat of the ball is 0.855 J/(g°C)

6. For a calorimeter, if a ball with a mass of 0.2 kg, has an initial temperature of 38 °C when placed in a calorimeter filled with water, what is the final temperature (in degrees. Celsius) of the ball if the ball absorbs 33 J from the water. Specific heat of the ball is 0.855 J/(g°C)

Chemistry
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
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**Exercise 6: Calorimeter Calculation** For a calorimeter, if a ball with a mass of 0.2 kg has an initial temperature of 38°C when placed in a calorimeter filled with water, what is the final temperature (in degrees Celsius) of the ball if the ball absorbs 33 J from the water? The specific heat of the ball is 0.855 J/(g°C). **Explanation:** In this question, we are looking to find the final temperature of a ball after it absorbs a certain amount of heat energy. The given data includes: - Mass of the ball: \(0.2 \, \text{kg}\) - Initial temperature of the ball: \(38 \, °\text{C}\) - Heat absorbed by the ball: \(33 \, \text{J}\) - Specific heat capacity of the ball: \(0.855 \, \text{J/(g°C)}\) The specific heat capacity, denoted as \(c\), is the amount of heat required to change the temperature of 1 gram of a substance by 1°C. The formula used to calculate the change in temperature when a certain amount of heat is absorbed is: \[ q = mc\Delta T \] Where: - \(q\) is the amount of heat absorbed (in joules) - \(m\) is the mass of the substance (in grams) - \(c\) is the specific heat capacity (in joules per grams per degree Celsius) - \(\Delta T\) is the change in temperature (in degrees Celsius) Rearranging the formula to solve for \(\Delta T\): \[ \Delta T = \frac{q}{mc} \] **Solution:** First, convert the mass from kilograms to grams: \[ 0.2 \, \text{kg} = 200 \, \text{g} \] Using the given values in the formula: \[ \Delta T = \frac{33 \, \text{J}}{200 \, \text{g} \times 0.855 \, \text{J/(g°C)}} \] Calculate \(\Delta T\): \[ \Delta T = \frac{33}{171} \approx 0.193 \, °\text{C} \] Since the ball absorbs heat, its temperature will increase by \(\Delta T\): \[ \text
Transcribed Image Text:**Exercise 6: Calorimeter Calculation** For a calorimeter, if a ball with a mass of 0.2 kg has an initial temperature of 38°C when placed in a calorimeter filled with water, what is the final temperature (in degrees Celsius) of the ball if the ball absorbs 33 J from the water? The specific heat of the ball is 0.855 J/(g°C). **Explanation:** In this question, we are looking to find the final temperature of a ball after it absorbs a certain amount of heat energy. The given data includes: - Mass of the ball: \(0.2 \, \text{kg}\) - Initial temperature of the ball: \(38 \, °\text{C}\) - Heat absorbed by the ball: \(33 \, \text{J}\) - Specific heat capacity of the ball: \(0.855 \, \text{J/(g°C)}\) The specific heat capacity, denoted as \(c\), is the amount of heat required to change the temperature of 1 gram of a substance by 1°C. The formula used to calculate the change in temperature when a certain amount of heat is absorbed is: \[ q = mc\Delta T \] Where: - \(q\) is the amount of heat absorbed (in joules) - \(m\) is the mass of the substance (in grams) - \(c\) is the specific heat capacity (in joules per grams per degree Celsius) - \(\Delta T\) is the change in temperature (in degrees Celsius) Rearranging the formula to solve for \(\Delta T\): \[ \Delta T = \frac{q}{mc} \] **Solution:** First, convert the mass from kilograms to grams: \[ 0.2 \, \text{kg} = 200 \, \text{g} \] Using the given values in the formula: \[ \Delta T = \frac{33 \, \text{J}}{200 \, \text{g} \times 0.855 \, \text{J/(g°C)}} \] Calculate \(\Delta T\): \[ \Delta T = \frac{33}{171} \approx 0.193 \, °\text{C} \] Since the ball absorbs heat, its temperature will increase by \(\Delta T\): \[ \text
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