The heat required to evaporate 0.15 kg of lead at 1750°C will be 130,650 J.
The movement of energy from a hot to a cold item is characterized as heat. Heat energy flows from a hot material to a cold one.
This occurs because faster-vibrating molecules transmit their energy to slower-vibrating ones.
The given data in the problem is;
m is the mass of lead = 0.15 kg
T is the temperature = 1750°C,
The latent heat of vaporization for lead is, [tex]\rm L_V[/tex] = 871 × 10³ J/kg.
The heat is found as;
[tex]\rm Q= m \times L_V \\\\ \rm Q= 0.15 \times 871 \times 10^3 \\\\ Q=130,650 \ J[/tex]
Hence the heat required to evaporate 0.15 kg of lead at 1750°C will be 130,650 J.
To learn more about the heat refer to the link;
brainly.com/question/1429452
