Answer to Question #344111 in Molecular Physics | Thermodynamics for rox

Molecular Physics | Thermodynamics

Question #344111

It takes 880 J to raise the temperature of 350 g of lead from 0°C to 20.0°C. What is the specific heat of lead?

Expert's answer

"c=\\frac Q{m\u2206t}=126~\\frac{J}{kg\\cdot \u00b0C}."

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