When 300 J of heat is added to 25 gm of sample of a material its temperature rises from 25∘Cto45∘C. the thermal capacity of the sample and specific heat of the material are respectively given by
A
15J/∘C,600J/kg∘C
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B
600J/∘C,15J/kg∘C
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C
150J/∘C,60J/kg∘C
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D
Noneofthese
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Solution
The correct option is A15J/∘C,600J/kg∘C Thermalcapacity=mc=QΔT=30045−25=30020=15J/∘CSpecificheat=ThermalcapacityMass=1525×10−3=600J/kg∘C