Antimatter doesn’t really do anything by it’s own, but if we let 1 kg react with 1 kg of matter (non-anti-matter), we get E = mc2 with m = 2 kg. So 1.8 * 1017 J, or 1.8 * 1011 MJ. If we assume that 10 MJ/kg is represented by about 1 cm, the bar would have to be 1.8 * 1010 cm or about 1.8 * 108 m. A standard A4 piece of paper is about 30 cm tall, so 6.0 * 108 A4 papers are needed. I.e. 600 million papers.
So we definitely have enough paper, but it would be a very tall stack.
Bah, that graph needs antimatter.
Is there enough paper on earth?
Antimatter doesn’t really do anything by it’s own, but if we let 1 kg react with 1 kg of matter (non-anti-matter), we get E = mc2 with m = 2 kg. So 1.8 * 1017 J, or 1.8 * 1011 MJ. If we assume that 10 MJ/kg is represented by about 1 cm, the bar would have to be 1.8 * 1010 cm or about 1.8 * 108 m. A standard A4 piece of paper is about 30 cm tall, so 6.0 * 108 A4 papers are needed. I.e. 600 million papers.
So we definitely have enough paper, but it would be a very tall stack.