Question

a machine is used to convert heat energy into work energy, and is connected to a cooling system which keeps the machine at a constant temperature. if 1000 kj of heat energy enter the machine, and the machine performs 760 kj of work, how much heat energy must the cooling system remove?

Explanation:

Step1: Use the first law of thermodynamics

The first law of thermodynamics is $\Delta U = Q - W$. Since the machine is at a constant temperature, $\Delta U=0$ (internal energy change is zero). So, $Q = W$. Here, $Q$ is the heat input, and part of it is used for work $W$ and part is removed as heat $Q_{out}$. So, $Q_{in}=W + Q_{out}$.

Step2: Rearrange the formula to find $Q_{out}$

We can rewrite the formula as $Q_{out}=Q_{in}-W$. Given $Q_{in} = 1000\space kJ$ and $W = 760\space kJ$.
Substitute the values: $Q_{out}=1000 - 760$.

Answer:

$240\space kJ$