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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?

Respuesta :

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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

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