3-5 g of a fuel having molecular weight 28 - was burnt in a caloriemeter and raised the temperature of 1 g of water from 25- C to 67-3- C- If all the heat generated was used in heating water- calculate the magnitude of molar heat of combustion of fuel in calmol -1-

Amount of fuel used = 3.5 g/28 gmol^ 1=0.125 mol So the energy generated, C_p(water)× increase in temperature× amount of water in g =1cal/g^oC× 42.30 ^oC× ...

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Byju's Answer

Standard XII

Chemistry

Difference between Internal Energy and Enthalpy

3.5 g of a fu...

Question

3.5 g of a fuel (having molecular weight 28), was burnt in a caloriemeter and raised the temperature of 1 g of water from 25 $^{\circ} C$ to 67.3 $^{\circ} C$ . If all the heat generated was used in heating water, calculate the magnitude of molar heat of combustion of fuel in $c a l m o l^{- 1}$ .

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Solution

Amount of fuel used = $\frac{3.5 g}{28 g m o l^{- 1}} = 0.125 m o l$
So the energy generated, $C_{p} (w a t e r) \times i n c r e a s e i n t e m p e r a t u r e \times a m o u n t o f w a t e r i n g$
$= 1 c a l / g^{o} C \times 42.30^{o} C \times 1 g = 42.3 c a l$
So, the molar heat of combustion, $\frac{42.3}{0.125} c a l m o l^{- 1} = 338.4 c a l m o l^{- 1}$

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3.5 g of a fuel (having molecular weight 28), was burnt in a caloriemeter and raised the temperature of 1 g of water from 25∘C to 67.3∘C. If all the heat generated was used in heating water, calculate the magnitude of molar heat of combustion of fuel in calmol−1.

Amount of fuel used = 3.5 g28 gmol−1=0.125 mol So the energy generated, Cp(water)×increase in temperature×amount of water in g =1cal/goC×42.30 oC×1 g=42.3 cal So, the molar heat of combustion, 42.30.125 calmol−1=338.4 calmol−1

3.5 g of a fuel (having molecular weight 28), was burnt in a caloriemeter and raised the temperature of 1 g of water from 25 $^{\circ} C$ to 67.3 $^{\circ} C$ . If all the heat generated was used in heating water, calculate the magnitude of molar heat of combustion of fuel in $c a l m o l^{- 1}$ .