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 ^oC)
∆T = change in temperature = Tfinal - Tinitial in ^oC

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