Question

Calculate δe, if the system absorbs 7.24 kj of heat from the surroundings while its volume remains constant (assume that only p−v work can be done). express your answer using three significant figures.

Answer 1

Answer:

$\delta Q = \delta E = 7.24 kJ$

Explanation:

Heat absorbed by the system is given as

$\delta Q = 7.24 kJ$

now from first law of thermodynamics we know that

$\delta Q = \delta E + W$

here

W = work done by the system

$\delta E$ = change in internal energy

also we know that when volume of the system remains same then work done by the system must be zero

$W = 0$

so from above equation

$\delta Q = \delta E = 7.24 kJ$

Answer 2

I believe the answer is 7.24 kJ.
From the equation ΔE = dW + dQ; where W is the work done on/by the system and Q is the heat the system absorbs/loses.
Therefore; ΔE = 72.4 kJ since the system has bot done any p-v work (dV= zero) and has absorbed heat.