A chunk of lead at 91.6 C was added to 200.0g of water at 15.5 C. The specific heat of lead is 0.129J/g C. When the temperature stabilized, the temperature of the mixture was 17.9 C. Assuming no heat was lost to the surroundings, what was the mass of lead added?

Question

Aliana Mckenzie

Chemistry

27 July, 12:29

A chunk of lead at 91.6 C was added to 200.0g of water at 15.5 C. The specific heat of lead is 0.129J/g C. When the temperature stabilized, the temperature of the mixture was 17.9 C. Assuming no heat was lost to the surroundings, what was the mass of lead added?

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Answers (1)

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Luciana · Answer 1

Let's apply the conservation of energy through the equation below:

Q for lead + Q for water = 0

So,

Q for lead = - Q for water

where

Q = mass*specific heat * (T₂ - T₁)

The specific heat of liquid water is 4.187 J/g·°C

Substituting the values:

(m) (0.129 J/g·°C) (17.9 - 91.6°C) = - (200 g) (4.187 J/g·°C) (17.9 - 15.5°C)

Solving for m,

m = 211.39 grams of lead