Question

A sample of hexane (C6H14) has a mass of 0.580 g. The sample is burned in a bomb calorimeter that has a mass of 1.900 kg and a specific heat of 3.21 J/giK. What amount of heat is produced during the combustion of hexane if the temperature of the calorimeter increases by 4.542 K?

Answer 1

Use the formula, Q= mcT

Q= heat
m= mass= 1.900Kg= 1.900 x 10^3 grams
c= specific heat= 3.21 
T= 4.542 K

Q= (1.900 x10^3g)(3.21)(4.542K)= 14.6 Joules.

Answer 2

Answer : The amount of heat produced during the combustion of hexane per mole is 4109.79 kJ/mole.

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

Formula used :

$Q=m\times c\times \Delta T$

where,

Q = heat absorb = ?

m = mass of calorimeter = 1.900 kg  = 1900 g

c = specific heat of calorimeter = $3.21J/g.K$

$\Delta T$ = change in temperature = 4.542 K

Now put all the given value in the above formula, we get:

$Q=1900g\times 3.21J/g.K\times 4.542K$

$Q=27701.658J=27.70kJ$

Now we have to calculate the moles of hexane.

$\text{Moles of }C_6H_{14}=\frac{Mass of }C_6H_{14}}{Molar mass of }C_6H_{14}=\frac{0.580g}{86.18g/mole}=0.00674mole$

Now we have to calculate the heat produced during the combustion of hexane per mole.

$\text{Heat produced}=\frac{27.70kJ}{0.00674mole}=4109.79kJ/mole$

Therefore, the amount of heat produced during the combustion of hexane per mole is 4109.79 kJ/mole.