Question

A 2.60 g lead weight, initially at 10.7 ∘C, is submerged in 8.18 g of water at 52.0 ∘C in an insulated container What is the final temperature of both the weight and the water at thermal equilibrium?

Answer #1

Q = m c ∆T

Q = quantity of heat in joules (J)

m = mass of the substance acting as the environment in

grams (g)

c = specific heat capacity (4.19 for H2O) in J/(g
^{o}C)

∆T = change in temperature = Tfinal - Tinitial in ^{o}C

Heat lost by water = heat gained by lead

specific heat capacity of lead = 0.16

specific heat capacity of water = 4.18

mass of lead = 2.6 gm

Mass of water = 8.18 gm

2.6 x 0.16 x (Tf-10.7) = 8.18 x 4.18 x (52-Tf)

0.416 Tf - 4.451 = 1778 - 34.19 Tf

34.6Tf = 1782.455

**Tf = 51.5 ∘C**

**final temperature of both the weight and the water at
thermal equilibrium is** **51.5
∘C**

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