If the heat of combustion of gasoline is 47 kj/g, how many grams of gasoline are required to heat 2500.0 ml of water in a bomb calorimeter from an initial temperature of 21.9°c to a final temperature of 67.8°c?

a. 14.0 g
b. 10.2 g
c. 2.44 g
d. no right answer.

10,2 grams of gasoline are required to heat 2500.0 mL of water in a bomb calorimeter (option b.)

Explanation:

1 g of gasoline has 47 kJ as heat of combustion, so let's calculate our heat with the data the excersie give us.

First of all we should know water density which is 1 g/ml, so we assume that 2500 mL are been occuped by 2500 g.

Q = m . C . ΔT

Q = 2500 g . 4,186 J / g ºC (Tfinal - Tinitial)

Q= 2500 g . 4,186 J / g ºC ( 67,8ºC - 21,9 ºC)

Q = 480343,5 J

Now, we convert the J to kJ to make the rule of three

480343,5 J / 1000 = 480,3 kJ

By the end:

47 kJ 1 g

480,3 kJ (480,3 kJ . 1g) / 47 kJ = 10,2 g

solute : a solute is a substance that is dissolved in another substance is known as a solvent. it is mixed with a special mixture know as homogeneous mixture.

solvent: a solvent is a substance that dissolved a solute, which results in a solution is known as solvent.

the transfer of heat by electromagnetic radiation is called thermal radiation . it is usually carried out by electromagnetic waves in an energy range just below the visible spectrum, which is the infrared range.

so the answer would be radiation.

: )