Chemistry, 26.11.2019 01:31 billlyyyyyyyyyy

Calculate the change internal energy (δe) for a system that is giving off 65.0 kj of heat and is performing 855 j of work on the surroundings. calculate the change internal energy (δe) for a system that is giving off 65.0 kj of heat and is performing 855 j of work on the surroundings. 9.00 x 102 kj 64.1 kj -9.00 x 102 kj -64.1 kj -65.9 kj

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A system that is giving off 65.0 kJ of heat and is performing 855 J of work on the surroundings has a change in the internal energy of -65.9 kJ.

The system is giving off 65.0 kJ of heat (Q). By convention, when a system is releasing heat, Q < 0. Then, Q = -65.0 kJ.The system is also performing 855 J of work (W) on the surroundings. By convention, when a system performs work on the surroundings, W < 0. Then, W = -855 J (-0.855 kJ)

According to the First Law of Thermodynamics, we can calculate the change in the internal energy (ΔE) of the system using the following expression.

$\Delta E = Q + W = (-65.0 kJ) + (-0.855 kJ) = -65.9 kJ$

A system that is giving off 65.0 kJ of heat and is performing 855 J of work on the surroundings has a change in the internal energy of -65.9 kJ.

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Answer from: shadowselena63

-64.1 kJ.

Explanation:

According to first law of thermodynamics:

$\Delta E=q+w$

$\Delta E$ =Change in internal energy

q = heat absorbed or released

w = work done or by the system

w = work done by the system= $-P\Delta V$ {Work is done by the system is negative as the final volume is greater than initial volume}

w = -855 Joules = 0.855 kJ (1kJ=1000J)

q = -65.0 kJ {Heat released by the system is negative}

$\Delta E=-65.0J-(-0.855J)=-64.1kJ$

Thus the change internal energy (ΔE) for a system that is giving off 65.0 kJ of heat and is performing 855 J of work on the surroundings is -64.1 kJ.

Answer from: Quest

answer: enthalpy

explanation:

for a exothermic reaction, the energy is released i.e. the energy of the products is lower than the energy of the reactants.

$\delta h=e_p-e_r$

$\delta h$ = enthaply

$e_p$ = energy of products

$e_r$ = energy of reactants

as $e_p< e_r$ , thus $\delta h$ comes out to be negative. negative value of enthalpy shows that the reaction is exothermic.

Answer from: Quest

because of all the things that happen around the us, many people get into arguments which sadly could get physical, but we should actually make a change for that.

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