Chemical engineer studying the properties of fuels placed 1.690 g of a hydrocarbon in the bomb of a calorimeter and filled it with o2 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 (qv) per gram of the fuel? (d for water = 1.00 g/ml; c for water = 4.184 j/g·°c.) enter your answer in scientific notation.

heat of combustion  is 23258.17 J/g or 23.258 kJ/g

Explanation:

given data

temperature range = 20.00°C to 23.55°C

so temperature change is = 3.55°C

heat capacity = 403 J/K

so here heat absorbed by calorimeter is

heat absorbed = heat capacity × change temperature

heat absorbed = 403 × 3.55

heat absorbed = 1430.65 J

and

volume of water is = 2.550 L = 2550 mL

and water density = 1.00 g/mL

so mass of water = volume × density

mass of water = 2550 × 1

mass of water = 2550 g

and

specific heat capacity for water is here = 4.184 J/g-°C

and temperature change = 3.55°C

so heat absorbed by water = mass × specific heat × temperature change

heat absorbed by water = 2550 × 4.184 × 3.55 = 37875.66 J

so

so heat absorbed by water and calorimeter  is 1430.65 J + 37875.66 J

heat absorbed by water and calorimeter  is 39306.31 J

so this heat give combustion 1.690 g fuel

so for 1 gram heat given =

heat of combustion  is 23258.17 J/g or 23.258 kJ/g

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.