connersmart53 connersmart53

15-05-2020
Chemistry

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3. How much heat, in joules, is needed to raise the temperature of 125.0 g of Lead (Clead=
0.130 J/gºC) from 17.5°C to 41.1°C?

Respuesta :

valeherrera333 valeherrera333

19-05-2020

Answer:

Q= 383.5 J

Explanation:

Q= m*C*dT

Q=(125.0 g)*(0.13 J/ g °C)* (41.1 °C-17.5°C)

Q= 383.5 J

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