The energy stored in chemical bonds is a form of kinetic thermal energy potentia

Question

The energy stored in chemical bonds is a form of kinetic thermal energy potential energy mechanical energy The first law of thermodynamics states that energy is transferred from the surroundings to the system during a combustion reaction if a system loses energy to the surroundings, then the surroundings must do an equal amount of work on the system if the surroundings gain energy from the system then the system must lose an equal amount of energy energy is transferred from the system to the if a system does work on the surroundings, then the surroundings must transfer an equal amount of energy to the system. What is the change in internal energy (delta E) of a system when 5.00 times 10^3 J of work is done on the system while it releases 10.0KJ of energy to the surroundings? -5.00 kJ - 10.0kJ + 15.0 kJ + 5010 J + 4990 J What is the change in internal energy (delta E) of a system when it is heated with 35.0 J of energy while it does 15 0 J of work? +50, 0 J -50.0 J -20.0 J +20.0 J +35.0 J Calculate the work performed on by a gas when it is compressed under a constant pressure of 8.00 a. atm from 95.0L to 9.5 L, (101.3 J = 1 L atm). + 77.0kJ + 69.3 kJ + 85.5 J - 85.56 J How much energy is needed to change the temperature of 25.00 mL of water from 10.0 degree C to 95.0 degree C? 160 times 10^2 282 2.kJ 8.89 kJ 6, 41kJ 105kJ A 1.20 kg piece of granite [c_p = 0.790 J/(g middot C)] at 60.0 degree C must beta cooled to 24.0 degree C by submersion in water. How much water [c_p = 4.18 J/(g middot C)] initially at 18 degree C would beta required?0.302 kg 0.907 kg 1.36 kg 1.72 kg 8.16 kg

Explanation / Answer

Answers:

1) c = Potential energy

Sometimes, it is equal to the chemical energy of the bonds.

2) C

3) c = +15 kJ

Given Data:

Q = 10 kJ

W = 5 kJ

It is work done on the system, so Del.E = Q + W = 10kJ + 5 kJ = +15 kJ

4) d = +20 J

Given Data:

Q = 35 J

W = 15 J

It is work done by the system, so Del.E = Q - W = 35 J - 15 J = +20 J

5) c = 69.3 kJ

Pressure, P = 8 atm

Vf, final volume after compression = 9.5 L

Vi, Initial volume = 95 L

Work, W = -P(Vf - Vi) = - 8 atm (9.5 - 95) = 684 atm. L * 101.3 J = 69.3 J

6) c =8.89 kJ

Givendata:

m = 25 mL * 1 g/mL / 18 g/mol = 1.39 moles

Cp = 75.38 J / mol. C

T1 = 10 ° C

T2 = 95 ° C

Q = m. cp . Del.T = 1.39 moles * 75.38 J / mol. C * (95 - 10) ° C = 8890 J = 8.89 kJ

7) C = 1.36 kg

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