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What is the change in heat when 35.0 g of H2O freezes?

A) -20.3 kJ
B) 20.3 kJ
C) -40.6 kJ
D) 40.6 kJ

1 Answer

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Final answer:

To calculate the heat change when 35.0 g of water freezes, we convert the mass to moles and multiply by the enthalpy of fusion, yielding a heat change of approximately -11.7 kJ, suggesting an error in the provided options.

Step-by-step explanation:

The student asked what the change in heat is when 35.0 g of H2O freezes. This is a chemistry question involving the concept of enthalpy change during a phase transition. To determine the change in heat when water freezes, we can use the enthalpy of fusion for water, which is -6.01 kJ/mol at 0 °C. Since freezing is an exothermic process, the heat change will be negative, indicating that heat is being released.

First, we need to convert the mass of water to moles:

• The molar mass of water (H2O) is approximately 18.01 g/mol.
• Moles of H2O = mass (in grams) / molar mass = 35.0 g / 18.01 g/mol = 1.944 moles.

Now, we multiply the moles of water by the enthalpy of fusion to find the total heat change:

• Heat change = moles × enthalpy of fusion = 1.944 moles × (-6.01 kJ/mol) ≈ -11.7 kJ.

The closest answer to our calculated value of -11.7 kJ is option A) -20.3 kJ, which suggests that none of the provided options exactly matches our calculation. There may be a typo or mistake in the question. If we assume that -11.7 kJ per one freeze-thaw cycle, then two cycles would indeed yield -23.4 kJ, which is closest to option A) -20.3 kJ.

User Aimee Jones
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