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
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A Thermalcapacity=mc=QΔT=30045−25=30020=15J/∘CSpecificheat=ThermalcapacityMass=1525×10−3=600J/kg∘C