Question

In a reaction, gaseous reactants form a liquid product. The heat absorbed by the surroundings is 1.1 MJ, and the work done on the system is 13.2 kcal. Calculate ΔE (in kJ). Be sure to include the correct sign (+/-). Enter to 0 decimal places.

Answer 1

Answer: $\Delta E$ is 1155 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 by the system=$-P\Delta V$  {Work done on the system is positive as the final volume is lesser than initial volume}

w =$13.2kcal=55.2kJ$   (1kcal=4.184 kJ)

q = +1.1 MJ = 1100 kJ  (1MJ=1000kJ)   {Heat absorbed by the system is positive}

$\Delta E=1100kJ+(55.2)kJ=1155kJ$

Thus $\Delta E$ is 1155 kJ