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zlopas [31]
3 years ago
15

A 32.8 g iron rod, initially at 22.4 C, is submerged into an unknown mass of water at 63.1 C, in an insulated container. The fin

al temperature of the mixture upon reaching thermal equilibrium is 59.1 C.What is the mass of the water?
Chemistry
1 answer:
laila [671]3 years ago
3 0

Answer:

mass water = 32.4 g

Explanation:

specific heat iron = 0.450 J/g°C

specific heat water = 4.18 J/g°C

32.8 x 0.450 ( 59.1 - 22.4) + mass water x 4.18 ( 59.1- 63.1)=0

541.7 - mass water x 16.7 = 0

mass water = 32.4 g

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The percent by mass sugar of a solution : 11.07%

<h3>Further explanation</h3>

Given

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mass of water = 100 g

Required

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Percent mass of sugar :

\tt \%mass=\dfrac{mass~sugar}{mass~solution}\times 100\%\\\\\%mass=\dfrac{12.45}{112.45}\times 100\%=11.07\%

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