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Question
1 kg of molten lead at its melting point of 327°C is dropped into 1 kg of water at 20°C. Assuming no loss of heat, calculate the final temp. of water. (Sp. heat of lead = 130 J/kg°C, latent heat of lead = 27000 J/kg and Sp. heat of water = 4200 J/kg°C).
Solution
Let the final temperature of the mixture be t
Heat lost by lead = Heat gained by water
m1L + m1 x CL x (327 - t) = mz x cw x (t - 20)
1 x 27000 + 1 x 130 x (327 - t) = 1 x 4200 x (t - 20)
27000 + 42510 _ 130t = 4200t - 84000
t = 35.45°C
So, the final temperature of water is 35.45°C.
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