Question
Asked Aug 27, 2019
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A 0.547 g0.547 g sample of steam at 104.3 °C104.3 °C is condensed into a container with 4.82 g4.82 g of water at 15.1 °C15.1 °C. What is the final temperature of the water mixture if no heat is lost? The specific heat of water is 4.18  J g⋅ °C4.18  J g⋅ °C, the specific heat of steam is 2.01  J g⋅ °C2.01  J g⋅ °C, and Δ?vap=−40.7 kJ/mol.

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Expert Answer

Step 1

The mixing of steam and water involves the loss of heat energy by steam which will be gained by cooler water. Hence, the expression for heat is written as:

Heat lost by steam +heat used to warm cooler water after mixing = heat gained by water after mixing

 

The expression for heat change associated with a system is given in the equation (1) in which Q is the heat change, m is the mass of substance, s is the specific heat of substance and ΔT is the change in temperature.

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Q msAT . (1)

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Step 2

The heat loss by steam is equal to the product of heat of vaporization per mole (ΔHvap) and the number of steam (n) and the expression for heat required to warm water and cool steam is written from the equation (1).

 

Hence, the final expression is written as:

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nAH (T-T) (T. T, 1(steam m vap (steam steam water ( water 1(water

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Step 3

Substitute the required values in the equation ...

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0.547 g -40700 J/mol - (4.82 g) (4.18 J/g°c) (T,-15.1)c 18.0 g/mol (0.547 g) (2.01 Jg°C) (T -104.3) °C -1236.83 J 1.09 J/oC(T)-113.69 J 20.15 J/C(T,)-304.23 J 19.06T 1045.5 T2 54.9 °C

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