A bomb calorimeter has a heat capacity of 675 J/°C and contains 925 g of water. If the combustion of 0.500 mole of a hydrocarbon increases the temperature of the calorimeter from 24.26°C to 53.88°C, determine the enthalpy change per mole of hydrocarbon. The specific

heat of water = 4.184 J g-1 °C-1.

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What is the enthalpy change when 175 g of C3H8 are burned in excess O2?

C3H8(g) + 5 O2(g) ® 3 CO2(g) + 4 H2O(l) DH° = -2220 kJ

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A sample of water containing 2.00 moles is initially at 30.0°C. If the sample absorbs 2.00 kJ of heat, what is the final temperature of the water? (specific heat of water = 4.184 J g-1 °C-1)

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heat of water = 4.184 J g-1 °C-1.

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What is the enthalpy change when 175 g of C3H8 are burned in excess O2?

C3H8(g) + 5 O2(g) ® 3 CO2(g) + 4 H2O(l) DH° = -2220 kJ

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A sample of water containing 2.00 moles is initially at 30.0°C. If the sample absorbs 2.00 kJ of heat, what is the final temperature of the water? (specific heat of water = 4.184 J g-1 °C-1)

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Temperature increase, dT = 53.88^{0}C – 24.26^{0}C = 29.62^{0}C

Heat capacity of Calorimeter = 675 J ^{0}C^{-1}

Specific heat of water = 4.186 J g^{-1}^{0}C^{-1}

Total heat released during combustion of hydrocarbon is absorbed by the calorimeter system (calorimeter itself + water in it).

So, heat gained by calorimeter system is equal to heat gained by calorimeter plus heat gained by water.

Note: Heat capacity is the amount of heat required to raise the temperature by 1^{0}C (or any other temperature unit). It does not specify mass of a object. Instead it tells that the object (here, calorimeter) as a while requires this amount of energy to increase temperature by 1^{0}C. So,

**Heat gained by calorimeter = (dT x heat capacity of calorimeter)**

Specific heat capacity or specific heat is the amount of heat required to raise the temperature of 1.0 g object by 1^{0}C. So,

Heaty gained by water = m.s.dT ; where, s= specific heat of water

Therefore, total heat gained by calorimeter system is-

Or, q = heat gained by calorimeter + heat gained by water

Or, q = (C x dT) _{calorimeter} + msdT _{water}

Or, q = (675 J ^{0}C^{-1} x 29.62^{0}C) + [925g x (4.186 J g^{-1}^{0}C^{-1}) x 29.62^{0}C]

Or, q = 17968.5 J + 24627.686 J = 42596.186 J

**Thus,** total heat gained by calorimeter system = 42596.186 J

Since, “Total heat released during combustion of hydrocarbon is absorbed by the calorimeter system (calorimeter itself + water in it)”,-

Heat released by 0.5 mol hydrocarbon = **42596.186 J**

Heat released per mol hydrocarbon = (42596.186 J/ 0.5 mol)

= **85192.372 J/ mol**

Thus, molar enthalpy of combustion of hydrocarbon = **85192.372 J/ mol**

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