Question

the balanced combustion reaction for c₆h₆ is
2c₆h₆(l) + 15o₂(g) → 12co₂(g) + 6h₂o(l) + 6542 kj
if 7.100 g c₆h₆ is burned and the heat produced from the burning is added to 5691 g of water at 21 °c, what is the final temperature of the water?
final temperature: °c

Explanation:

Step1: Calculate the moles of \(C_6H_6\)

The molar mass of \(C_6H_6\) (\(M\)): \(M=(6\times12.01 + 6\times1.008)\space g/mol=78.108\space g/mol\)
The number of moles of \(C_6H_6\) (\(n\)): \(n=\frac{m}{M}\), where \(m = 7.100\space g\)
\(n=\frac{7.100\space g}{78.108\space g/mol}\approx0.0909\space mol\)

Step2: Calculate the heat released (\(q\))

From the balanced equation \(2C_6H_6(l)+15O_2(g)\to12CO_2(g)+6H_2O(l)+6542\space kJ\), for \(2\space mol\) of \(C_6H_6\) burned, the heat released is \(6542\space kJ\)
For \(n = 0.0909\space mol\) of \(C_6H_6\), \(q=\frac{0.0909\space mol}{2\space mol}\times6542\space kJ\)
\(q=\frac{0.0909\times6542}{2}\space kJ\approx296.7\space kJ = 296700\space J\)

Step3: Use the heat - transfer formula \(q = mc\Delta T\)

The specific heat capacity of water \(c = 4.184\space J/(g\cdot^{\circ}C)\), \(m = 5691\space g\), \(q=296700\space J\), and \(\Delta T=T_f - T_i\), \(T_i = 21^{\circ}C\)
\(\Delta T=\frac{q}{mc}\)
\(\Delta T=\frac{296700\space J}{5691\space g\times4.184\space J/(g\cdot^{\circ}C)}\)
\(\Delta T=\frac{296700}{5691\times4.184}\space^{\circ}C\approx12.3^{\circ}C\)

Step4: Calculate the final temperature (\(T_f\))

\(T_f=T_i+\Delta T\)
\(T_f=21^{\circ}C + 12.3^{\circ}C=33.3^{\circ}C\)

Answer:

\(33.3\)