Calculate ∆E for the following case:

q = +1.62 kJ and w = -874 J

a. -872 J
b. -746 J
c. +746 J
d. +872 J
e. +2.49 kJ

The internal energy(ΔE) of a substance is calculated below:

From the first law of thermodynamics;

ΔE=q+w

Explanation:

ΔE=q+w

here;

+q=endothermic reaction

-q=exothermic reaction

+w=work done on the system

-w=work done by the system

Given:

q=+1.62kJ=1620J

w=-874J

To solve:

the internal energy(ΔE)

We know:

ΔE=q+w

according to the problem;

ΔE=q-w

since;

w=-874J (i.e.)work is done by the system.

ΔE=1620-874

ΔE=+746J

Therefore the internal energy is +746J

i.e. the option is "c"(+746J)