A system releases 130 kJ of heat while 109 kJ of work is done on the system. Calculate the change in internal energy

Before we write the given data let us first mention some sign convention that we use when we solve the thermodynamics sums 

Work is done by the system it is taken as POSITIVE

Work is done on the system it is taken as NEGATIVE

Energy entering a system is taken as POSITIVE

Energy leaving a system is taken as NEGATIVE

Given data - 

dQ = -130 kJ

dW = -109 kJ

To find, change in internal energy dU =?

Solution -

By first law of thermodynamics we have,

dQ = dU + dW

dU = dQ - dW

dU = -130 - (-109) = -130 + 109 = -21 kJ

So there is a decrease in the internal energy by 21kJ