Question

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

Answer 1

`604000\ \text{J}`

Step-by-step explanation:

m = Mass of ice = 200 g

`\Delta T` = Temperature change of water =
`(100-0)^(\circ)\text{C}`

c = Specific heat capacity of water = 4200 J/kg °C

`L_f` = Specific latent heat of fusion = 340 kJ/kg

`L_v` = Specific latent heat of vaporisation = 2260 kJ/kg

Heat required to convert ice to water =
`mL_f`

Heat required to raise the temperature of water to boiling point =
`mc\Delta T`

Heat required to convert water to steam =
`mL_v`

Total heat required

`q=mL_f+mc\Delta T+mL_v\\\Rightarrow q=m(L_f+c\Delta T+L_v)\\\Rightarrow q=0.2(340* 10^3+4200(100-0)+2260* 10^3)\\\Rightarrow q=604000\ \text{J}`

Heat required to convert the given amount of ice to steam at the required temperature is
`604000\ \text{J}`.