A 175 g lump of molten lead at its melting point (327 C) is placed into 55.0 g of water at 20.0 C. The specific heat of lead is 130.J/kg C and the Hf of lead is 20,400 J/kg. When the lead and the water have reached equilibrium, what is the temperature of the mixture?

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Annainn Annainn · Answer 1 · 2019-02-11T14:22:45+01:00

Ans: Temperature of mixture = 61.8 C

Step 1: Find the temperature of water when lead is placed in contact

Q = mcΔT

ΔT = Q/mc = .175 kg*20400 J.Kg-1/.055 kg * 4187 J.kg-1C-1 = 15.5 C

T₂-T₁ = 15.5

T₂ = 20.0 + 15.5 = 35.5 C

Step 2: Find the temperature of mixture at equilibrium

Heat lost by lead = heat gained by water

-[mc(T2-T1)]lead = [mc(T2-T1)]water

-(0.175*130(T2-327)) = (0.055*4187(T2-35.3)

T2 = 61.8 C

onyebuchinnaji onyebuchinnaji · Answer 2 · 2022-03-31T12:35:53+02:00

The temperature of the mixture when they reach equilibrium is 61.8 ⁰C.

The temperature of water when lead is placed in contact is calculated by using heat capacity formula as shown below;

Q = mcΔT

ΔT = Q/mc

ΔT = (0.175 x 20400)/(0.055 kg x 4200) = 15.5 ⁰C

ΔT = T₂-T₁

T₂-T₁ = 15.5

T₂ = 20.0 + 15.5 = 35.5 C

The temperature of the mixture at equilibrium is determined by applying the principle of conservation of energy.

Heat lost by lead = heat gained by water

mc(T1 - T2)]lead = [mc(T2-T1)]water

0.175*130(327 - T2)) = (0.055*4187(T2-35.3)

T2 = 61.8 C

Thus, the temperature of the mixture when they reach equilibrium is 61.8 ⁰C.

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