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?

1
Expert's answer
2022-05-23T15:18:54-0400

c=Qm∆t=126 Jkg⋅°C.c=\frac Q{m∆t}=126~\frac{J}{kg\cdot °C}.


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Canada #334539 on Jan 2023