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How much heat would need to be removed to cool 50.3 g of water from 25.6°C to -10.7°C?

How much heat would need to be removed to cool 50.3 g of water from 25.6°C to -10.7°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
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 12 of 35** **Thermal Properties of a Substance:** 1. **C\_solid** = 2.092 J/g°C - This is the specific heat capacity of the substance in its solid state. 2. **C\_liquid** = 4.184 J/g°C - This is the specific heat capacity of the substance in its liquid state. 3. **T\_fusion** = 0.0°C - This is the temperature at which the substance transitions from solid to liquid, known as the fusion temperature. 4. **ΔH\_fusion** = 6.01 kJ/mol - This is the enthalpy change required for the substance to change from solid to liquid at its fusion temperature. **Interactive Calculator:** - A numerical keypad is provided for input, allowing calculations involving heat capacities and fusion enthalpies. - Tap or pull up for additional resources for further learning and exploration of these concepts.
Transcribed Image Text:**Question 12 of 35** **Thermal Properties of a Substance:** 1. **C\_solid** = 2.092 J/g°C - This is the specific heat capacity of the substance in its solid state. 2. **C\_liquid** = 4.184 J/g°C - This is the specific heat capacity of the substance in its liquid state. 3. **T\_fusion** = 0.0°C - This is the temperature at which the substance transitions from solid to liquid, known as the fusion temperature. 4. **ΔH\_fusion** = 6.01 kJ/mol - This is the enthalpy change required for the substance to change from solid to liquid at its fusion temperature. **Interactive Calculator:** - A numerical keypad is provided for input, allowing calculations involving heat capacities and fusion enthalpies. - Tap or pull up for additional resources for further learning and exploration of these concepts.
**Question 12 of 35:** **Topic:** Heat Removal in Cooling Water **Question:** How much heat would need to be removed to cool 50.3 g of water from 25.6°C to -10.7°C? **Given Data:** - Specific heat capacity of solid (\( C_{\text{solid}} \)): 2.092 J/g°C - Specific heat capacity of liquid (\( C_{\text{liquid}} \)): 4.184 J/g°C - Fusion temperature (\( T_{\text{fusion}} \)): 0.0°C - Enthalpy of fusion (\( \Delta H_{\text{fusion}} \)): 0.341 kJ/g **Diagram:** This section contains input boxes for entering calculations, along with a numerical keypad and options for mathematical operations. **Instructions:** Determine the total heat to be removed by considering the steps of cooling liquid water, freezing at the fusion point, and further cooling the resulting ice. Use the provided constants in calculations. Additionally, you can refer to additional resources by tapping on the option provided.
Transcribed Image Text:**Question 12 of 35:** **Topic:** Heat Removal in Cooling Water **Question:** How much heat would need to be removed to cool 50.3 g of water from 25.6°C to -10.7°C? **Given Data:** - Specific heat capacity of solid (\( C_{\text{solid}} \)): 2.092 J/g°C - Specific heat capacity of liquid (\( C_{\text{liquid}} \)): 4.184 J/g°C - Fusion temperature (\( T_{\text{fusion}} \)): 0.0°C - Enthalpy of fusion (\( \Delta H_{\text{fusion}} \)): 0.341 kJ/g **Diagram:** This section contains input boxes for entering calculations, along with a numerical keypad and options for mathematical operations. **Instructions:** Determine the total heat to be removed by considering the steps of cooling liquid water, freezing at the fusion point, and further cooling the resulting ice. Use the provided constants in calculations. Additionally, you can refer to additional resources by tapping on the option provided.
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