Calorie calculation
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If one wants to use water to cool ferric tetraoxide, reducing its temperature from 470 degrees to 70 degrees, then should the temperature of the water rise to 70 degrees as well (assuming room temperature to be 20 degrees)? How can one calculate how much water is needed for this cooling process? The weight of ferric tetraoxide is calculated as 1 ton.
It is necessary to know the specific heats of ferric tetraoxide and water, and then calculate using energy conservation – the heat released by ferric tetraoxide is equal to the heat absorbed by water
I did the calculations as well, using the heat fusion formula, that is, Cpm*b*△t. For water, I used its boiling point of 100 degrees and calculated based on 20 liters changing to 70 degrees. However, the result I got was too low. To cool down iron, 1.7 tons of water are needed for every ton of iron. I’m not sure where the error lies
In practical applications, operability must be taken into account; it depends on the method used for cooling. 1. Is it acceptable for the water used for cooling to vaporize? 2. There is also the issue of the efficiency of water-based cooling – when the temperature difference is too small, is the cooling efficiency sufficient? 3. Is it feasible to cool 1 ton of material using 1.7 tons of water?
It’s a theoretical calculation at the moment, but I haven’t worked it out yet; I keep feeling that the calculation is incorrect – does it really take 1.7 tons of water to cool 1 ton of iron by 400 degrees? That’s too unlikely. Water is allowed to vaporize. As for efficiency, it hasn’t been considered yet; the basic usage amounts are still not determined
Have you considered a two-step calculation? The first step is to determine the amount of water needed to cool the iron to 100 degrees, taking into account both the latent heat and the sensible heat of water. For the second step, let’s calculate: it takes water to cool iron from 100 degrees to 70 degrees, and here the latent heat of water is not needed. If the calculation results are as you said, that’s normal too, as the sensible heat and latent heat of water are notoriously high.
I just calculate it in two steps; 100 degrees is the dividing line. The final result is around 1.7 tons, but I just find this figure a bit hard to believe. Could anyone help me calculate precisely how much water is needed for cooling?
The water doesn’t necessarily have to be at 70 degrees, right? If it were at 70 degrees, there would be no issue with latent heat of vaporization. What method of cooling do you plan to use?
Just do the heat balance calculation: W = Cp * M * Δt
70 degrees is the temperature to which we intend to cool the iron, so calculations are done based on this temperature. The method of cooling involves simply placing the iron inside to cool it down. Iron is cooled from 470 degrees; it will definitely vaporize. Why isn’t the latent heat of vaporization used in the calculations?