Question

Calculate the heat required to convert 2kg of ice at -12°c to steam at 100°c. L for ice= 336000j/kg, L for steam= 2260000. C for ice = 2100j/kg/k

Answer 1

Heat required to heat the ice from -12^oC to the melting (0^oC):

$Q_1=c_{ice}m\Delta{T}=2100J/kg\cdot{K}\times2kg\times(0\degree{C}-(-12\degree{C}))=50400J$

Heat required to melt the ice at 0^oC:

$Q_2=L_{ice}m=336000J/kg\times2kg=672000J$

Heat required to heat the water from 0^oC to the boiling (100^oC):

$Q_3=c_{water}m\Delta{T}=4184J/kg\cdot{K}\times2kg\times(100\degree{C}-0\degree{C})=836800J$

Heat required to evaporate the water at 100^oC:

$Q_4=L_{steam}m=2260000J/kg\times2kg=4520000J$

The total heat is the sum of the heats required for each step:

$Q=Q_1+Q_2+Q_3+Q_4=50400J+672000J+836800J+4520000J=6080000J=6080kJ$

Answer: 6080 kJ