Question

1 kg of molten lead at its melting point of 327°C is dropped in 1 kg of water at 20°C. Assuming no heat is lost; calculate the final temperature of water. (Sp. heat cap. of lead = 130 J/kg°C, Sp. latent heat of fusion of lead = 27000 J/g, Sp. heat cap. of water = 4200 j/kg°C)

Answer 1

Let the final temperature of the mixture be t.

Heat lost by lead = Heat gained by water 

m₁L + M₁ x C_L x (327 - t) = m₂ x C_w x (t - 20)

1 x 27000 + 1 x 130 x (327 - t) = 1 x 4200 x (t - 20)

27000 + 42510 - 130t = 4200t - 84000

153510 = 4330 t

t = 35.45°C

So, the final temperature of water is 35 .45°C.