A sample of ethanol (C2H6O) has a mass of
0.2301 g. Complete combustion of this sample causes the temperature of a bomb calorimeter to increase by 1.33°C. The calorimeter has a mass of 2.000 kg and a specific heat of 2.45 J/g•°C.

What is the heat of combustion for this sample?

-6.517 kJ is the heat of combustion for this sample that is 0.2301 grams of ethanol.

Explanation:

Mass of bomb calorimeter = 2.000 kg = 2.000 × 1000 g = 2000. g

1 kg = 1000 g

Specific heat of bomb  calorimeter = c = 2.45 J/g°C

Change in temperature =

Heat absorbed by the bomb claorimeter = Q

Heat energy released on combustion of 0.2301 grams of ethanol = Q'= -Q

Q' = -Q = -6,517 J = -6.517 kJ

1 J = 0.001 kJ

-6.517 kJ is the heat of combustion for this sample that is 0.2301 grams of ethanol.

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

i think the answer is terrae but not 100 percent sure. ▓▬▬ §