Calculate the heat required to convert 3 kg of ice at `-12^(@)C` kept in a calorimeter to steam at `100^(@)C` at atmospheric pressure. Given,
specific heat capacity of ice = `2100 J kg^(-1) K^(-1)`
specific heat capicity of water = `4186 J kg^(-1)K^(-1)`
Latent heat of fusion of ice = ` 3.35 xx 10^(5) J kg^(-1)`
and latent heat of steam = ` 2.256 xx 10^(6) J kg^(-1)` .

Answer

9.1

Explanation

Mass of the ice,
m= 3 kg
Specific heat capacity of ice,
`c_("ice")=2100 J kg^(-1)K^(-1)`
Specific heat capacity of water,
`c_("water")=4186 J kg^(-1)K^(-1)`
Latent heat of fusion of ice,
`(L_f)_("ice")=3.35 xx 10^5 J kg^(-1)`
Latent heat of steam,
`L_("steam")=2.256 xx 10^6 J kg^(-1)`
Heat required to convert ice at `-12^@C` to ice at `0^@C`
`Q_1=mc_("ice")DeltaT_1`
`=(3kg)(2100 J kg^(-1)K^(-1))[0-(-12)]^@C=75600J`
Heat required to melt ice at `0^@C` to water at `0^@C`
`Q_2=m(L_f)_("ice")=(3kg)(3.35 xx 10^5 J" "kg^(-1))`
=1005000 J
Heat required to convert water at `0^@C` to water at `100^@C`,
`Q_3=mc_w Delta T_2=(3kg)(4186" J kg"^(-1)K^(-1))(100^@C)`=1255800 J
Heat required to convert water at `100^@C` to steam at `100^@C`,
`Q=mL_("steam")=(3kg)(2.256 xx 10^6 J kg^(-1))=6768000` J
Total heat required to convert 3kg of ice at `-12^@C` to steam at `100^@C`,
`Q=Q_1+Q_2+Q_3+Q_4`
=75600 J + 1005000 J + 1255800 J + 6768000 J
`=9.1 xx 10^6` J