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Calculate the total energy transferred when 200 g of ice cubes at 0°C are changed to steam at 100°C.

Specific heat capacity of water = 4200 J/kg °C
Specific latent heat of fusion = 340 kJ/kg
Specific latent heat of vaporisation = 2260 kJ/kg

Sagot :

boffeemadrid

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].

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