Question

1. a system underwent a cyclic process using path 1 and path 2, in path 1, it did 200j work, but in path 2 it got 300j work and released 400j heat, please calculate the heat in path 1.

Explanation:

Step1: Recall the first law for cyclic process

For a cyclic process, the change in internal energy \(\Delta U = 0\) because the system returns to its initial state. The first law of thermodynamics is \(\Delta U=Q - W\) (where \(Q\) is heat added to the system and \(W\) is work done by the system). For a cyclic process, \(\Delta U = 0\), so the net heat \(Q_{net}\) and net work \(W_{net}\) must satisfy \(Q_{net}=W_{net}\).

Let's define the work and heat for each path:
- For path 1: Let \(Q_1\) be the heat (we need to find it) and \(W_1 = 200\space J\) (work done by the system, so positive as per sign convention where work done by system is positive).
- For path 2: The system got work, so work done on the system, so \(W_2=- 300\space J\) (since work done by system is negative when work is done on it). The system released heat, so heat is removed from the system, \(Q_2=-400\space J\) (heat added to system is positive, so released heat is negative).

Step2: Calculate net work and net heat

Net work \(W_{net}=W_1 + W_2=200+( - 300)=- 100\space J\)

Net heat \(Q_{net}=Q_1 + Q_2=Q_1+( - 400)\)

Since for cyclic process \(Q_{net}=W_{net}\) (because \(\Delta U = 0\) and \(\Delta U=Q - W\) implies \(Q = W\) for cyclic process), we have:

\(Q_1-400=- 100\)

Step3: Solve for \(Q_1\)

Add 400 to both sides of the equation:

\(Q_1=- 100 + 400=300\space J\)

Wait, but let's check the sign convention again. Maybe the sign convention is different: some use \(\Delta U = W+Q\) where \(W\) is work done on the system. Let's re - define:

If \(W\) is work done on the system, then for path 1: work done by system is 200J, so work done on system \(W_1=- 200\space J\)

For path 2: work done on system \(W_2 = 300\space J\)

Heat released by system is 400J, so heat added to system \(Q_2=-400\space J\)

For cyclic process \(\Delta U = 0\), and \(\Delta U=W + Q\) (where \(W\) is work done on system, \(Q\) is heat added to system)

So \(0=(W_1 + W_2)+(Q_1 + Q_2)\)

Substitute values: \(0=(-200 + 300)+(Q_1-400)\)

Simplify: \(0 = 100+Q_1 - 400\)

\(0=Q_1 - 300\)

So \(Q_1 = 300\space J\)? Wait, no, if heat is added to the system, positive. But in path 1, if the system did work (200J), and in path 2, work was done on the system (300J) and heat was released (400J). Let's use the correct sign convention from thermodynamics:

The first law is \(\Delta U=Q - W\), where:

- \(Q\): positive if heat is added to the system, negative if heat is removed.

- \(W\): positive if work is done by the system, negative if work is done on the system.

For a cyclic process, \(\Delta U = 0\), so \(Q_{net}=W_{net}\)

Path 1: \(W_1 = 200\space J\) (work done by system, positive), \(Q_1=\)?

Path 2: \(W_2=-300\space J\) (work done on system, so \(W\) is negative), \(Q_2=-400\space J\) (heat removed from system, so \(Q\) is negative)

\(W_{net}=W_1 + W_2=200+( - 300)=- 100\space J\)

\(Q_{net}=Q_1 + Q_2=Q_1+( - 400)\)

Since \(Q_{net}=W_{net}\)

\(Q_1-400=- 100\)

\(Q_1=400 - 100 = 300\space J\)? Wait, no, that would mean heat is added to the system in path 1. But let's think again.

Wait, the net work done by the system: in path 1, system does 200J work (so \(W_1 = 200\)). In path 2, system gets 300J work, which means work is done on the system, so the work done by the system is \(W_2=- 300\). So net work done by system \(W_{net}=200-300=- 100\space J\) (negative means net work is done on the system).

Net heat: in path 2, system releases 400J, so \(Q_2=-400\) (heat out of system). Let \(Q_1\) be heat in path 1 (heat into system is positive, heat out is negative). Then net heat \(Q_{net}=Q_1+Q_2=Q_1 - 400\)

For cyclic process, \(\Delta U = 0\), so \(Q_{net}=W_{net}\) (because \(\Delta U=Q - W\) and \(\Delta U = 0\) implies \(Q = W\))

So \(Q_1-400=- 100\)

\(Q_1=400 - 100=300\space J\). Wait, but that would mean heat is added to the system in path 1. But let's check with another approach.

The change in internal energy for the cycle is zero. So the sum of \(Q - W\) over the cycle is zero.

For path 1: \(\Delta U_1=Q_1 - W_1\)

For path 2: \(\Delta U_2=Q_2 - W_2\)

Since \(\Delta U_1+\Delta U_2 = 0\) (cyclic process), we have \((Q_1 - W_1)+(Q_2 - W_2)=0\)

\(Q_1+Q_2=W_1 + W_2\)

We know \(W_1 = 200\space J\) (work done by system), \(W_2=-300\space J\) (work done on system, so \(W_2\) is negative when work done by system is considered), \(Q_2=-400\space J\) (heat released by system, so \(Q_2\) is negative)

So \(Q_1+( - 400)=200+( - 300)\)

\(Q_1-400=- 100\)

\(Q_1=300\space J\)

Wait, but if the system did work in path 1 (200J) and in path 2, work was done on it (300J) and heat was released (400J), then the heat in path 1 should be 300J? But let's check the signs again. Maybe the problem uses \(W\) as work done on the system. Let's re - define:

Let \(W\) be work done on the system. Then:

Path 1: system does 200J work, so work done on system \(W_1=-200\space J\)

Path 2: system gets 300J work, so work done on system \(W_2 = 300\space J\)

Heat: path 2, system releases 400J, so heat added to system \(Q_2=-400\space J\)

For cyclic process, \(\Delta U = 0\), and \(\Delta U=Q + W\) (where \(Q\) is heat added to system, \(W\) is work done on system)

So \(0=(Q_1 + Q_2)+(W_1 + W_2)\)

Substitute values: \(0=(Q_1-400)+(-200 + 300)\)

\(0=Q_1-400 + 100\)

\(0=Q_1-300\)

\(Q_1 = 300\space J\)

So regardless of the sign convention (as long as we are consistent), we get \(Q_1 = 300\space J\). Wait, but maybe I made a mistake. Let's think about the net work and net heat.

Net work done by the system: in path 1, system does 200J (positive), in path 2, system has work done on it, so work done by system is - 300J. Net work \(W_{net}=200-300=-100\space J\) (system has net work done on it).

Net heat: in path 2, system releases 400J (so heat out, \(Q_2=-400\)), in path 1, let \(Q_1\) be heat in (positive) or out (negative). For cyclic process, \(\Delta U = 0\), so \(Q_{net}=W_{net}\) (because \(\Delta U=Q - W\) and \(\Delta U = 0\) implies \(Q = W\)). So \(Q_1+Q_2=W_1 + W_2\)

\(Q_1-400=200-300\)

\(Q_1-400=-100\)

\(Q_1=300\space J\)

So the heat in path 1 is 300J (heat is added to the system in path 1).

Answer:

The heat in path 1 is \(\boldsymbol{300\space J}\) (heat is added to the system).