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.