Question

What is the change in internal energy (in J) of a system that absorbs 0.297 kJ of heat from its surroundings and has 0.750 kcal of work done on it? Give your answer in scientific notation. × 10,

Answer 1

According to the first law of thermodynamics, the change in internal energy ($\Delta$E) is the sum of heat (q) and work (w):

$\Delta$E = q + w

Since heat is absorbed by the system from the surroundings, heat (q) is positive; also, work (w) is positive since work is done on the system.

Converting the values of heat to work to Joules, we get:

q = 0.297 kJ $\times$ 10³ J $\div$ 1 kJ [1 kJ = 10³ J]

= 297 J

w = 0.750 kcal $\times$ 10³cal $\div$ 1 kcal $\times$ 4.184J $\div$ 1 cal [1 kcal = 10³cal ; 1 cal = 4.184 J]

= 3138 J

Therefore, $\Delta$E_int = q + w = +297 J - [-3138 J] = 3435 J

There is an increase in internal energy.

3435 J can be written as :

Change in internal energy = 3.435 $\times$ 10³ J