10 grams of ice at 0 degree centigrade absorbs 5460 joules of heat to melt and change itno water at 50 degree centigrade.calculate the specific latent heat of fusion of ice?
heat=heatmeltingIce+heatRisingTempWater
5640=10*Lf + 10*cwater*(50-0)
put in the specific heat of water cw, in joules/gram-C, and solve for heat of fusion, Lf.
To calculate the specific latent heat of fusion of ice, we need to use the equation:
Q = m × L
Where:
Q is the heat energy absorbed or released during the phase change
m is the mass of the substance
L is the specific latent heat of fusion
In this case, the ice absorbs 5460 joules of heat to melt, so we can write the equation as:
5460 J = 10 g × L
First, we need to convert the mass from grams to kilograms, since the unit of specific latent heat is usually expressed in J/kg. The conversion factor is 1 g = 0.001 kg. So, the mass of ice is 10 g × 0.001 kg/g = 0.01 kg.
Now, we can substitute the mass back into the equation:
5460 J = 0.01 kg × L
To solve for L, we divide both sides of the equation by 0.01 kg:
L = 5460 J / 0.01 kg
L ≈ 546,000 J/kg
Therefore, the specific latent heat of fusion of ice is approximately 546,000 J/kg.