Question #168510

When 732.5 J of heat are added to 102G of a solid substance the temperature of the solid increases by 48.8 K what is the specific heat capacity of the solid


1
Expert's answer
2021-03-08T06:11:41-0500

Q=mcΔTc=QmΔT=732.5102×48.8=0.147  J/gKQ = mcΔT \\ c = \frac{Q}{mΔT} \\ = \frac{732.5}{102 \times 48.8} \\ = 0.147\;J/gK

Answer: 0.147 J/gK


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