Question

You finally decide to take some liquid nitrogen from the physics lab and freeze the 1.5 pounds of beans which were sitting in your vegetable basket at 20 c, rather than cook and eat them as your mother would have you do. beans are mostly water, like all living things. if the specific heat of water is approximately 1 cal/g/c and latent heat of fusion of water is approximately 80 cal/g, and the latent heat of vaporization of liquid nitrogen is 45 cal/g as measured by you, how much liquid nitrogen should you take from the lab (approximately)?you don't want to take a whole lot more than you really need.

Answer 1

So in this problem, we treat the beans as if it's water. From 20°C, we need to bring it down to 0°C at freezing point. The total heat needed for that transformation would be

Q = m(Cp,waterΔT + ΔHfus) = (1.5 lb)(1 kg/2.2 lb)(1000g/1 kg)[(1 cal/g·°C)(20 - 0°C) + 80 cal/g)
Q = 68181.82 cal

Now, we have to equate this to the heat for liquid nitrogen.

Q = mΔHvap
68181.82 cal = m(45 cal/g)
Solving for m,
m = 1,515.15 g or 3.33 pounds of liquid nitrogen