Calculate the number of - kJ- necessary to raise the temperature of - 60-0 g- -160- of -160- aluminum from - 35-o C- to - 55-o C- Molar heat capacity of -160- - Al- is -160- 24 Jmol - -1 K - -1-

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Standard XII

Chemistry

Molar and Specific Heat Capacity

Calculate the...

Question

Calculate the number of $k J$ necessary to raise the temperature of $60.0 g$ of aluminum from $35^{o} C$ to $55^{o} C$ . Molar heat capacity of $A l$ is ${24 J m o l}^{- 1} K^{- 1}$ .

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Solution

Given: $T_{1} = 35^{o} C = (35 + 273) K = 308 K$
$T_{2} = 55^{o} C = (55 + 273) K = 328 K$
$C_{p} = 24 J m o l^{- 1} K^{- 1}$
Number of mole of Aluminium $= \frac{60 g}{27 g / m o l} = 2.23 m o l e$
Using the relation,
$△ H = n C_{p} △ T$
$△ H = 2.23 m o l e \times 24 J m o l^{- 1} K^{- 1} \times 20 K$
$△ H = 1.07 K J$

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Calculate the number of kJ necessary to raise the temperature of 60.0 g of aluminum from 35o C to 55o C. Molar heat capacity of Al is 24 Jmol−1K−1.

Given: T1=35oC=(35+273)K=308 KT2=55oC=(55+273)K=328 KCp=24 Jmol−1K−1Number of mole of Aluminium =60 g27 g/mol=2.23 moleUsing the relation, △H=nCp△T△H=2.23 mole×24 Jmol−1K−1×20 K△H=1.07 KJ

Calculate the number of $k J$ necessary to raise the temperature of $60.0 g$ of aluminum from $35^{o} C$ to $55^{o} C$ . Molar heat capacity of $A l$ is ${24 J m o l}^{- 1} K^{- 1}$ .