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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?​

User Jfcogato
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1 Answer

3 votes

Answer:

Q= 383.5 J

Step-by-step explanation:

Q= m*C*dT

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

Q= 383.5 J

User SPKoder
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3.9k points