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

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.