latent heat of fusion formula
Latent heat of fusion is the amount of heat received by a solid item to transform it into a liquid without increasing the temperature further. Because sea ice and brine may live together at any temperature and melt at a temperature other than oC when bathed in a concentrated salt solution, the amount of latent heat is complicated in the case of sea ice, much as it is in the walls of brine cells when brine cells migrate.
The heat absorbed by the substance, or the latent heat of the fusion formula, is represented as when m kg of solid changes to a fluid at a fixed temperature, which is its melting point. L = m * L denotes the substance’s particular latent heat of fusion.
The heat that the material absorbs or releases is represented as Q = mc t as the temperature of the substance varies from t1 (low temperature) to t2 (high temperature).
mc (t2 – t1) = Q
Q = mL + mc Δt is the total quantity of heat absorbed or emitted by the substance.
1. At 20°C, a piece of metal weighs 60g. 0.5g of the steam condenses on it when it is submerged in a steam stream at 100°C. Given that the latent heat of steam is 540 cal/g, calculate the specific heat of the metal.
Let c be the metal’s specific heat.
The metal has gathered heat.
Q = m c Δt
Q = 60 × c × (100 – 20)
Q = 60 × c × 80 cal
The steam’s heat is expelled.
Q = m × L
Q = 0.5 × 540 cal
According to the concept of mixing,
The amount of heat provided is equal to the amount of heat absorbed.
0.5 × 540 = 60 × c ×80
c = 0.056 cal/g °C
2. If 64500 calories of heat are extracted from 100 g of steam at 100 degrees C, calculate the amount of water transformed to ice. The latent heat of ice and steam, respectively, is 80 cal per gram and 540 cal per gram.
The temperature of the water will drop from 100 degrees C to 10 degrees C if the complete steam is transformed into water at 100 degrees C. Also, a portion of the water at 0 degrees C for ice formation.
The heat required to convert steam at 100 degrees Celsius to water at 100 degrees Celsius = mLsteam = 100540cal = 54000 calories.
The water content is 6.25 grams.