How much heat (in kJ) would need to be removed to cool 113.1 g of water from 25.60°C to -10.70°C? Cliquid = 4.184 J/g•°C Csolid = 2.092 J/g.°C AHtusion = Trusion = 0.0 °C 6.01 kJ/mol

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Chapter1: Chemical Foundations
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**Question:**

How much heat (in kJ) would need to be removed to cool 113.1 g of water from 25.60°C to -10.70°C?

**Given Data:**

- **C(solid)** = 2.092 J/g°C
- **C(liquid)** = 4.184 J/g°C
- **T(fusion)** = 0.0°C
- **ΔH(fusion)** = 6.01 kJ/mol

**Explanation:**

In this problem, you are tasked with calculating the amount of heat removed to cool a given mass of water through its phase transition from liquid to solid and further cooling below freezing. The specific heat capacities for both solid and liquid phases are provided, along with the enthalpy of fusion and the temperature at which fusion occurs. The process involves cooling the liquid water to 0°C, freezing it, and then further cooling the solid ice to the final temperature.
Transcribed Image Text:**Question:** How much heat (in kJ) would need to be removed to cool 113.1 g of water from 25.60°C to -10.70°C? **Given Data:** - **C(solid)** = 2.092 J/g°C - **C(liquid)** = 4.184 J/g°C - **T(fusion)** = 0.0°C - **ΔH(fusion)** = 6.01 kJ/mol **Explanation:** In this problem, you are tasked with calculating the amount of heat removed to cool a given mass of water through its phase transition from liquid to solid and further cooling below freezing. The specific heat capacities for both solid and liquid phases are provided, along with the enthalpy of fusion and the temperature at which fusion occurs. The process involves cooling the liquid water to 0°C, freezing it, and then further cooling the solid ice to the final temperature.
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