A 2.94 g lead weight, initially at 11.1 ∘C, is submerged in 7.72 g of water at 51.8 ∘C in an insulated container. What is the final temperature of both the weight and the water at thermal equilibrium?
m(lead) = 2.94 g
T(lead) = 11.1 oC
C(lead) = 0.128 J/goC
m(water) = 7.72 g
T(water) = 51.8 oC
C(water) = 4.184 J/goC
T = to be calculated
Let the final temperature be T oC
use:
heat lost by water = heat gained by lead
m(water)*C(water)*(T(water)-T) = m(lead)*C(lead)*(T-T(lead))
7.72*4.184*(51.8-T) = 2.94*0.128*(T-11.1)
32.3005*(51.8-T) = 0.3763*(T-11.1)
1673.1649 - 32.3005*T = 0.3763*T - 4.1772
T= 51.3 oC
Answer: 51.3 oC
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