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A sample of lead released 2,744 J when it cooled from 90 to 19 °C. What was the
mass of the sample of lead? (The specific heat of lead is 0.129 J/g x °C)

Respuesta :

DrMantisToboggan

Answer:

299.60 g

Explanation:

Use the formula q = mCΔT (q= heat, m= mass, c = specific heat for that specific substance, ΔT is change in temperature Final Temp - Initial Temp)

Released heat means it has negative sign.

q = mCΔT

-2744 = (mass) (0.129) (19-90)

mass= 299.60 g

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