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 final temperature of the mixture upon reaching thermal equilibrium is 59.1 C. What is the mass of the water?

Question

Sterling

Chemistry

15 November, 17:06

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 final temperature of the mixture upon reaching thermal equilibrium is 59.1 C. What is the mass of the water?

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Answers (1)

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Brett Green · Answer 1

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