Question

What is the change in internal energy (in J) of a system that does 4.50 ✕ 105 J of work while 3.20 ✕ 106 J of heat transfer occurs into the system, and 6.00 ✕ 106 J of heat transfer occurs to the environment?

Answer 1

Answer:

$-3.25\times 10^6 J$

Explanation:

We are given that

Work done by the system=$4.5\times 10^5$ J

Heat transfer into the system=$U_1=3.2\times 10^6$ J

Heat transfer to the environment=$U_2=6\times 10^6$ J

We have to find the change in internal energy

By first law of thermodynamics

$\Delta Q=\Delta U+w$

$\Delta Q=U_1-U_2=3.2\times 10^6-6\times 10^6=-2.8\times 10^6J$

Substitute the values then we get

$-2.8\times 10^6=\Delta U+4.5\times 10^5$

$\Delta U=-2.8\times 10^6-4.5\times 10^5=-28\times 10^5-4.5\times 10^5=-32.5\times 10^5=-3.25\times 10^6 J$

Hence, the change in internal energy =$-3.25\times 10^6 J$