Answer:
[tex]604000\ \text{J}[/tex]
Explanation:
m = Mass of ice = 200 g
[tex]\Delta T[/tex] = Temperature change of water = [tex](100-0)^{\circ}\text{C}[/tex]
c = Specific heat capacity of water = 4200 J/kg °C
[tex]L_f[/tex] = Specific latent heat of fusion = 340 kJ/kg
[tex]L_v[/tex] = Specific latent heat of vaporisation = 2260 kJ/kg
Heat required to convert ice to water = [tex]mL_f[/tex]
Heat required to raise the temperature of water to boiling point = [tex]mc\Delta T[/tex]
Heat required to convert water to steam = [tex]mL_v[/tex]
Total heat required
[tex]q=mL_f+mc\Delta T+mL_v\\\Rightarrow q=m(L_f+c\Delta T+L_v)\\\Rightarrow q=0.2(340\times 10^3+4200(100-0)+2260\times 10^3)\\\Rightarrow q=604000\ \text{J}[/tex]
Heat required to convert the given amount of ice to steam at the required temperature is [tex]604000\ \text{J}[/tex].
