Let's start by analyzing the compressor. The low pressure compressor. First law of the compressor says that 1w2 equals h1 minus h2. The lower case letters here, this is all specific quantities. Cp times t1 minus t2 because it's an ideal gas and we're assuming constant Cp. So that is going to be, now we know t1 but we don't know t2. We know about state 2 is p2 because we know the pressure ratio for the first and pressure stage. We also know that the first compressor stage isentropic. So, isentropic means t2 over t1 is equal to p2 over p1 to the power of gamma minus 1 over gamma. So t2 is 300 times 3 to the power of 0.4 over 1.4. And we get t2. Now for the intercooler. 2q3 is h3 minus h2 equals cp times t3 minus t2. We assume an ideal intercooler. So throughout our calculation we're being optimistic about the effect of intercooling. The best intercooling we can do, the most cooling we can do, is to get the gas down to ambient temperature. That is t1. So we'd have t3 equals t1 equals 300 carbon. So that gives us a 2q3 of cp is a thousand five. That's the cp for air, 300 carbon times t3, sorry, t2, no. What do we want? We want t3 minus t2. So, is minus 111.2 kilojoules per kilogram. I forgot to complete the compressor calculation. So, the, that's the work done by the compressor. The, that's the work done by, strictly, that's the work done by the gas in the compressor. It's negative, which reminds us that actually we're doing work on the gas in the compressor. And the compressor work is exactly equal to the intercooler heat transfer, which with hindsight is not surprising. Now, let's move on to the second stage compressor. We're doing pretty much the same way as the first. We're going to need to use the fact that it's isotropic compression. P4 over P3, we know is 10 over 3 equals t4 over t3. Well, to power of gamma minus 1 over gamma. Equals t4 over 3. So, we got t4 of...