Question
Asked Oct 25, 2019
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I am stuck on this problem and could use some guidance.

Thanks much. 

Exactly 14.4 mL of water at 33.0 °C is added to a hot iron skillet. All of the water is converted into steam at
100.0°C. The mass of the pan is 1.45 kg and the molar heat capacity of iron is 25.19 J/(molC). What is the
temperature change of the skillet?
°C
56.70
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Exactly 14.4 mL of water at 33.0 °C is added to a hot iron skillet. All of the water is converted into steam at 100.0°C. The mass of the pan is 1.45 kg and the molar heat capacity of iron is 25.19 J/(molC). What is the temperature change of the skillet? °C 56.70

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Expert Answer

Step 1

Given,

Specific heat capacity of water = 4.184J/g Celsius

Latent heat of vaporization of water = 2260J/g

Mass of iron pan = 1.45kg

No of moles in iron pan = 25.89 mol

Specific heat of iron = 25.19 J/mol℃

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volume of water given=14.4ml Mass of water =VolumexDensity 14.4 mlx1g/ml 14.4g

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Step 2

Initial temperature of water = 33℃

Heat abso...

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+mL water mcAT =ncAT water iron 14.4gx4.184J/g°C (100-33)°C+14.4g 2260J/g=25.89molx25.19J/molPCx(100-Q) 37183.2192J=652.169 (100-Q) 57.014-100-Q Q 42.98

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Chemical Thermodynamics

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