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50 g of metal piece at 27^(@)C requires ...

50 g of metal piece at `27^(@)C` requires 2400 J of heat energy so as to attain a temperature of `327^(@)C`. Calculate the specific heat capacity of the metal.

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Given, `m=50gor50/1000kg`
Q = 2400 J
`T_(1)=27^(@)C`
`T_(2)=327^(@)C`
We know, `Q=mcDeltat`
or `c=Q/(mDeltat)`
= `2400/(50/100xx(327-27))" "(becauseDeltat""^(@)C=DeltatK)`
= `2400/(5/100xx300)`
= `2400/15=160Jkg^(-1)K^(-1)`
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