3 ms·
Yes, you need to know the densities, but the models outputted the correct densities even as part of the wrong answer. Not sure what you mean by volumes "not ad
by realharo 2y ago
Yes, you need to know the densities, but the models outputted the correct densities even as part of the wrong answer.
Not sure what you mean by volumes "not adding". One way to calculate it is like:
density_alcohol = 0.789g/ml
density_water = 1g/ml
weight_total = 16g
(density = weight / volume)
volume_alcohol = 0.4 * volume_total
volume_water = 0.6 * volume_total
weight_total = weight_alcohol + weight_water
weight_alcohol =
density_alcohol * volume_alcohol =
0.789 * volume_alcohol =
0.789 * (0.4 * volume_total) =
0.3156 * volume_total
weight_water =
density_water * volume_water =
volume_water =
0.6 * volume_total
weight_total =
0.3156 * volume_total + 0.6 * volume_total =
0.9156 * volume_total
volume_total =
weight_total / 0.9156 =
16 / 0.9156 =
17.4749 (rounded)
volume_alcohol =
0.4 * volume_total =
6.99 ml (rounded)
Not sure which school grade problems like this are taught in, but I remember similar ones from waaay back. This particular one is a real world problem caused by having a kitchen scale but not a measuring cup, and me wrongly assuming that just asking ChatGPT would help. (ChatGPT 4 from about 2 months ago)
- Davidzheng 2y agohttps://www.flinnsci.com/api/library/Download/bc2d91b9d50f4ead80c9ff320b9a4419 https://www.flinnsci.com/api/library/Download/bc2d91b9d50f4e... I don't really know what ABV even means when the total volume of a mixture is not the sum of the alcohol volume and the water volume. I think practically it means your calculation is slight overestimation?
- realharo 2y agoOh, I had no idea! I didn't account for that (or a bunch of other things like temperature). So yeah, might be a bit off, but not as much as the wrong initial answers from the LLM - which also failed to mention this.