A175 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?

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

answer: a ball

explanation: because it will bounce