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?
1 answer:
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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