Answer:
12432 cal.
Explanation:
The process to change ice at -32 ºC to steam at 182 ºC can be divided into 5 steps:
1. Heat the ice to 0 ºC, which is the fusion temperature.
2. Melt the ice (obtaining liquid water), which is a process at constant pressure and temperature, so the liquid obtained is also at 0ºC.
3. Heat the liquid water from 0 ºC to 100 ºC, which is the vaporization normal temperature of the water.
4. Vaporization of all the water; this is also a process that occurs at constant pressure and temperature, so the produced steam will be at 100ºC.
5. Heat the steam from 100 ºC to 182 ºC.
Each process has a required energy, and the sum of the energy required for each and all of the steps is the total amount of energy required for the whole process:
[tex]E_T=E_1+E_2+E_3+E_4+E_5[/tex]
[tex]E_1[/tex] is a heating process for the ice, so we know that the energy required is proportional to the temperature difference through the specific heat:
[tex]E_1=m*Cp_{sol}*(T_2-T_1)\\E_1=16g*0.5\frac{cal}{gC}*(0-(-32))=256cal[/tex]
[tex]E_2[/tex] is a phase change process, so we do not use the specific heat (sensible heat), but the fusion heat (latent heat), so:
[tex]E_{2}=m*dh_{f}={16g*80\frac{cal}{g}}=1280cal[/tex]
Analogously,
[tex]E_3=m*Cp_{liq}*(T_3-T_2)=16g*1.00\frac{cal}{gC}*(100-0)C = 1600 cal[/tex]
[tex]E_{4}=m*{dh_{vap}}\\\\E_4=16g*540\frac{cal}{g} =8640cal[/tex]
[tex]E_{5}=m*Cp_{vap}*(T_{5}-T_{4})\\E_{5}={16g*0.5\frac{cal}{gK}*(182-100)K}=656cal[/tex]
Finally, the total energy required is:
[tex]E_T=256cal+1280cal+1600cal+8640cal+656cal\\E_T=12432cal[/tex]
