calculate the no.of KJ of heat necessory to raaise the temp. of 60g of aluminium from 35 C TO 55 C MOLAR HEAT CAPACITY OF ALUMINIUM IS 24J/MOL./K MOLAR MASS OF aluminium is 27g /mol.

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Solution

Initial temperature=35^oC
Final temperature=55^oC
ΔT=20^oC
molar heat capacity=24 Jmol^-1K^-1
$Q = m \times s \times Δ T Q = h e a t r e q u i r e d m = m a s s s = s p e c i f i c h e a t Δ T = c h a n g e i n t e m p e r a t u r e M o l a r h e a t c a p a c i t y i s g i v e n = 24 J m o l^{- 1} K^{- 1} S p e c i f i c h e a t = \frac{m o l a r h e a t c a p a c i t y}{m o l a r m a s s} H e n c e, Q = 60 \times \frac{24}{27} \times 20 = 1066.67 J = 1.067 K J$

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calculate the no.of KJ of heat necessory to raaise the temp. of 60g of aluminium from 35 C TO 55 C MOLAR HEAT CAPACITY OF ALUMINIUM IS 24J/MOL./K MOLAR MASS OF aluminium is 27g /mol.

Initial temperature=35oC Final temperature=55oC ΔT=20oC molar heat capacity=24 Jmol-1K-1 Q=m×s×ΔTQ=heat requiredm=masss=specific heatΔT=change in temperatureMolar heat capacity is given=24 J mol-1K-1Specific heat=molar heat capacitymolar massHence,Q=60×2427×20=1066.67J=1.067KJ