Question

If a system has 475 kcal of work done to it, and releases 5.00

Answer 1

475 kcal = 1987.4 kJ
of which 500 kJ are released as heat to the surroundings,
so the change in internal energy is +1487 kJ.

Answer 2

Q +w = U

-5*10^2 +275*4.184= U

U = 650.6 KJ

Answer 3

Work is done to th system.

Hence, it is positive quantity.

Heat is released to the surrounding.

Hence, it is negative quantity.

The expression for the change in the internal energy is $\Delta E = q +W$

Substitute values in the above expression.

$\Delta E = -5.00 \times 10^2 \quad kJ +475 \quad kcal \times \frac {4.184 \quad kJ}{1 \quad kcal}=1.49 \times 10^3 \quad kJ$