Question

when a 0.740 - g sample of trinitrotoluene (tnt), ( c_{7}h_{5}n_{2}o_{6} ), is burned in a bomb calorimeter, the temperature increases from ( 23.4^{circ}c ) to ( 26.9^{circ}c ). the heat capacity of the calorimeter is ( 534j/^{circ}c ), and it contains 675 ml of water. how much heat was produced by the combustion of the tnt sample? give answer in kj.

kj

the addition of 3.15 g of ( ba(oh)_{2}cdot 8h_{2}o ) to a solution of 1.52 g of ( nh_{4}scn ) in 100 g of water in a calorimeter caused the temperature to fall by ( 3.1^{circ}c ). assuming the specific heat of the solution and products is ( 4.20j/g^{circ}c ), calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation:

( ba(oh)_{2}cdot 8h_{2}o(s)+2nh_{4}scn(aq)\to ba(scn)_{2}(aq)+2nh_{3}(aq)+10h_{2}o(l) )

kj

Explanation:

Step1: Calculate the heat absorbed by the calorimeter and water for the TNT problem

The formula for heat is \(q = C\Delta T+mc\Delta T\).
For the calorimeter, \(C = 534\space J/^{\circ}C\), \(\Delta T=(26.9 - 23.4)^{\circ}C = 3.5^{\circ}C\). For water, \(m = 675\space g\) (since density of water is \(1\space g/mL\)), \(c = 4.184\space J/g^{\circ}C\), \(\Delta T = 3.5^{\circ}C\).
\(q_1=534\times3.5+675\times4.184\times3.5\)
\(q_1 = 1869+675\times14.644\)
\(q_1=1869 + 9884.7\)
\(q_1=11753.7\space J\)

Step2: Convert to kJ

Since \(1\space kJ=1000\space J\), \(q_1=\frac{11753.7}{1000}=11.8\space kJ\)

Step3: Calculate the heat absorbed for the \(Ba(OH)_2\cdot8H_2O\) problem

The total mass of the solution \(m=(3.15 + 1.52+100)\space g=104.67\space g\), \(c = 4.20\space J/g^{\circ}C\), \(\Delta T = 3.1^{\circ}C\)
Using \(q=mc\Delta T\)
\(q_2=104.67\times4.20\times3.1\)
\(q_2 = 104.67\times13.02\)
\(q_2=1363.8\space J\)

Step4: Convert to kJ

\(q_2=\frac{1363.8}{1000}=1.36\space kJ\)

Answer:

11.8 kJ, 1.36 kJ