Question

A system absorb 300 cal of heat with the result of that, the volume of the system becomes double of the initial volume and temperature changes from 273K to 546K. the work done by the system on the surroundings is 200.0 cal. Calculate $\Delta E$:

Answer 1

According to the thermodynamics first law, the energy can neither be created nor destroyed but can be transformed from one from to the other form, change in internal energy of a system is related to its heat and work done as follows.

$\Delta E$=$Q+W$

where

$\Delta E$ is change in internal energy of the system.

Q is a heat absorbed by the system=$+300cal$

W is the work done by the system on the surroundings =$-200$

Calculating the change in internal energy of the system from done using the equation

$\Delta E$=$Q+W$

$\Delta E$=$300cal+(-200cal)$

$\Delta =100cal$

Therefore the change in internal energy of the system is 100 cal.