A 32.9 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 62.6 ∘C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 58.4 ∘C.
What is the mass of the water?
Express your answer to two significant figures and include the appropriate units.
Let us denote iron by symbol 1 and water by symbol 2
m1 = 32.9 g
T1 = 22.4 oC
C1 = 0.45 J/goC
m2 = to be calculated
T2 = 62.6 oC
C2 = 4.184 J/goC
T = 58.4 oC
we have below equation to be used:
heat lost by 2 = heat gained by 1
m2*C2*(T2-T) = m1*C1*(T-T1)
m2*4.184*(62.6-58.4) = 32.9*0.45*(58.4-22.4)
m2*17.5728 = 532.98
m2= 30.3298 g
Answer: 30.3 g
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