A 31.4 g iron rod, initially at 22.3 ∘C, is submerged into an unknown mass of...
A 31.4 g iron rod, initially at 22.3 ∘C, is submerged into an unknown mass of water at 63.6 ∘C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 60.0 ∘C. What is the mass of the water?
Homework Answers
m(iron) = 31.4 g
T(iron) = 22.3 oC
C(iron) = 0.45 J/goC
m(water) = to be calculated
C(water) = 4.184 J/goC
T = 60.0 oC
We will be using heat conservation equation
use:
heat lost by water = heat gained by iron
m(water)*C(water)*(T(water)-T) = m(iron)*C(iron)*(T-T(iron))
m(water)*4.184*(63.6-60.0) = 31.4*0.45*(60.0-22.3)
m(water)*15.0624 = 532.701
m(water)= 35.3663 g
Answer: 35.4 g
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