Calculating Heat of Fusion and Capacity for Water: 50g Ice to 22°C Liquid Water

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SUMMARY

The calculation of heat required to convert 50g of ice at 0°C to liquid water at 22°C involves the heat of fusion and the heat capacity of water. The heat of fusion for water is 6.01 kJ/mol, and the heat capacity of liquid water is 75.3 J/mol·K. The correct approach requires multiplying the heat of fusion by the number of moles of ice, which is 2.778 mol, resulting in 16.7 kJ from the heat of fusion alone. Adding the heat required for the temperature change yields a total of 21.3 kJ, confirming the study guide's answer.

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Homework Statement


the heat of fusion of water is 6.01 kj/mol. the heat capacity of liquid water is 75.3 j/mol*k. the conversion of 50g of ice at 0 degrees C to liquid water at 22 degrees C requires _____kj of heat.

The Attempt at a Solution


i converted grams of ice to mols and got 2.778
then i multiplied by the heat capacity of 75.3 and the change in temperature of 22
i then converted 4601.667j to 4.601kj and added the heat of fusion 6.01kj/mol
i got an answer of 10.611kj but my study guide says the answer should be 21.3kj and I am not sure where in my calculations i went wrong...could someone please help?
 
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Take note that the heat of fusion is given in units of energy per mole i.e. you need to multiply the 6.01 by the number of moles.
 

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