Question

What is ΔE in kJ for a system that receives 1.79 kJ of heat from surroundings and has 4.51 kcal of work done on it at the same time. 1 cal = 4.184 J.

Answer 1

Answer: 20.7 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 on the system=$-P\Delta V$  {Work is done on the system is positive as the final volume is lesser than initial volume}

w = 4.51 kcal = $4.51\times 4.184kJ=18.9kJ$    (1kcal = 4.184kJ)

q = +1.79 kJ   {Heat absorbed by the system is positive}

$\Delta E=+1.79+(18.9)=20.7kJ$

Thus $\Delta E$ for a system that receives 1.79 kJ of heat from surroundings and has 4.51 kcal of work done on it at the same time is 20.7 kJ