36.8 kj
(Explanation): use latent heat of fusion value to solve. we will be using Q=(m)(Hfus) to solve this. Q=(1.5)(24.5) = 36.75, or 36.8. We don't need to convert kg to grams for this one because the units of Hfus are already in kg.
Strictly speaking, lead won't melt at 10 °C. You could do a theoretical calculation however by calculating the enthalpy change to take solid lead at 10 °C up its melting point at 327.5 °C, add the enthalpy of melting, and then subtract the enthalpy to cool liquid lead back to 10 °C. This last measurement would be only theoretical since lead will not remain liquid as you cool it below its melting point.
Enthalpy of melting = 22.4 kJ/kg
Specific heat of solid lead is ~130 J/kg/°C (its actually a function of temperature, but this average should suffice)
Specific heat of liquid lead is ~152 J/kg/°C (note that this is only theoretical below the melting point of lead.
Using these values, the theoretical energy required to melt 1 kg of lead at 10 °C would be:
130(327.5-10)J + 22400 J + 152(10-327.5)J = 15415 j = 15.415 kJ
36.8kJ
1935
No energy is gained. On the contrary, energy is required to make ice melt.
1,277,800 j
1935
200
1935
36.8 kj
mesons don't melt
heat of fusion
125.6 kJ (APEX)
No energy is gained. On the contrary, energy is required to make ice melt.
Energy is given off when cooling from a liquid to a solid. How much depends on the elements involved and the amount.
1,277,800 j
414 kJ
1935
1935
200
Iodine sublimes - it does not melt