Question

Heat of 30 kcal is supplied to a system and 4200j of external wok is done on the system so that it's volume decreases at constant pressure .What is the work done in its internal energy. (j=4200j/kcal)

Answer 1

SOLUTION :

Given:

Q=30 kcal Q=30 kcal is the heat added to the system (which makes it positive)W=4200 JW=4200 J is the work done ON the system (which is also positive)

Let us first convert our heat to Joules:

Q=30 kcal(4200 J1 kcal)Q=30 kcal(4200 J1 kcal)

we cancel the kcal units:

Q=30 kcal(4200 J1 kcal)Q=30 kcal(4200 J1 kcal)

we get:

Q=126,000 JQ=126,000 J

Since the heat was added to the system and the work was done on the system, then this means that both terms are positive. The change in internal energy is thus:

ΔU=126,000 J+4,200 JΔU=126,000 J+4,200 J

we thus get:

ΔU=130,200 J ≈ 130.2 kJΔU=130,200 J ≈ 130.2 kJ

Answer 2

Solution :-

 

$\bold{Δ Q \: = \: 30×4200}$

$\bold{=126000 \: J}$

$\bold{ΔW= −4200J}$

$\bold{∴Δu=ΔQ−ΔW}$

$\bold{=126000+4200J}$

$\bold{=1.302×105J}$