**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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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

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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