In order to answer this question, you need the enthalpy of fusion of ice, which is 333.55 J/g (Joules/gram). The enthalpy of fusion is the amount of heat that must be absorbed or lost in order to change physical state.
The number of Joules required to melt the ice = enthalpy of fusion of ice x mass of ice.
1 kg = 1000g
40kg x (1000g/1kg) = 40,000g
Joules needed to melt the ice = 333.55 J/g x 40000 g = 13342000 J or
1.3342 x 107 J
167.2 joules
334 j/g =167000 j
heat of fusion
Heat of fusion.
The energy required to melt a substance. (Apex)
226,ooo j
This heat is 32,48 joules.
To answer this, you need to know the ∆Hfusion of water, which happens to be 334 J/g. So, to melt 12.8 g of ice at 0ºC, the joules needed = (12.8 g)(334 J/g) = 4275 joules
Heat required to melt 1 g of ice at 0°C is approximately 80 cal . This is also called latent heat of fusion of ice.
In order to answer this question, you need the enthalpy of fusion of ice, which is 333.55 J/g (Joules/gram). The enthalpy of fusion is the amount of heat that must be absorbed or lost in order to change physical state. The number of Joules required to melt the ice = enthalpy of fusion of ice x mass of ice. Joules needed to melt the ice = 333.55 J/g x 40 g = 13342 J
334 j/g =167000 j
334 j/g =167000 j
The necessary heat is 9,22 joules.
heat of fusion
2,26 Kj are necessary
The energy required to melt a substance
The specific latent heat/ latent heat of fusion of ice is 333.55 J/gtherefore by using the equation E=mLE = 54 * 333.55= 18011.7 JTherefore the energy needed to melt 54 grams of ice at its melting point which is keeping it at 0 degrees Celsius is 18011.7 Joules.
The unit for the heat of vaporization is J/g or Joules per gram of material. The amount of energy required = mass * heat of vaporization = 5.0 g * 5.0 J/g = 25.0 J (Q.E.D.)