? newconnect.mheducation.com ork - Part 1 6 Saved 3 atemgts ertC Check my work B

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? newconnect.mheducation.com ork - Part 1 6 Saved 3 atemgts ertC Check my work Be sure to answer all parts. A chemical engineer studying the properties of fuels placed 1.130 g of a hydrocarbon in the bomb of a calorimeter and filled it with 02 gas. The bomb was immersed in 2.550 L of water and the reaction initiated. The water temperature rose from 20.00°C to 23.55°C. If the calorimeter (excluding the water) had a heat capacity of 403 J/K, what was the heat of reaction for combustion (gy) per gram of the fuel? (d for water 1.00 g/mL; c for water-4.184 J/gC.) Enter your answer in scientifie notation. x10Jg KPrev 8 of 1 Next 10

Explanation / Answer

Given that,

Temperature changes from 20.00 C to 23.55 C So, ?T = 3.55 C

Heat absorbed by calorimeter = heat capacity x temperature change

                                             = 403 J/C x 3.55

                                             = 1430.65 Joules

Volume of water = 2.55 L = 2550 mL

density of water = 1 g / mL

Now, mass of water = volume x density

                                 =2550 x 1

                                 =2550g

Specific heat capacity = 4.18 J/ gC

?T = 3.55 C

Heat absorbed by water = mass x specific heat x temperaturechange

                                      =2550 x 4.18 x 3.55

                                      =37839.45 Joules

Hence total heat absorbed =1430.65 Joules + 37839.45 Joules

=39270.1 Joules.

This is the heat given out by combustion of 1.130 g of fuel.

So for 1 g of fuel =39270.1 / 1.130

=34752.3 Joules

Hence heat of combustion = 34752.3 J/ g or 34.752 kJ/g

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