Question

505g of solid lead at 100 C is heated to its melting point and converted completely to liquid lead. Calculate the amount of thermal energy required for this to happen.

Answer 1

$m=505 \ g= 0.505 \ kg$

$T_i=100^{\circ} C$

$T_f=327.5^{\circ}C$

$c=130 \ J(^{\circ}C \ kg)$

$L_{f}=23\times 10^3 \ J/kg$

$Q \ - ?$

Solution:

At first, we need to increase the temperature of the lead from 100ºC to 327.5ºC (melting point).

$Q_1=cm(T_{f}-T_{i})=cm\Delta T$

$Q_1=130\times 0.505\times 227.5=14935.375 \ J$

Heat necessary for melting the lead:

$Q_2=L_{f} m$

$Q_2=23\times 10^3\times 0.505=11615 \ J$

Total heat required for this to happen:

$Q=Q_1+Q_2=26550.375\ J=26.5 \ kJ$

Answer: $26.5 \ kJ.$