Answer: 4276.2 calories Explanation: Given mass of steam is 6 gm at [tex]100^{\circ}C[/tex] Conversion of steam to ice involves steam to water at [tex]100^{\circ}C[/tex] water at [tex]100^{\circ}C[/tex] to water at [tex]0^{\circ}C[/tex] water to the ice at [tex]0^{\circ}C[/tex] Calories released during the conversion of steam to water at [tex]100^{\circ}C[/tex] [tex]E_1=mL_v\quad [L_v=\text{latent heat of vaporisation}]\\E_1=6\times 533=3198\ cal.[/tex] Calories released during the conversion of water at [tex]100^{\circ}C[/tex] to water at [tex]0^{\circ}C[/tex] [tex]E_2=mc(\Delta T)\quad [c=\text{specific heat of water,}1\ cal./gm.^{\circ}C]\\E_2=6\times 1\times 100=600\ cal.[/tex] Calories released during the conversion of water to the ice at [tex]0^{\circ}C[/tex] [tex]E_3=mL_f\quad [L_f=\text{latent heat of fusion}]\\E_3=6\times 79.7=478.2\ cal.[/tex] The total energy released is [tex]E=E_1+E_2+E_3\\E=3198+600+478.2=4276.2\ cal.[/tex] Log in to Reply
Answer: 4276.2 calories
Explanation:
Given
mass of steam is 6 gm at [tex]100^{\circ}C[/tex]
Conversion of steam to ice involves
Calories released during the conversion of steam to water at [tex]100^{\circ}C[/tex]
[tex]E_1=mL_v\quad [L_v=\text{latent heat of vaporisation}]\\E_1=6\times 533=3198\ cal.[/tex]
Calories released during the conversion of water at [tex]100^{\circ}C[/tex] to water at [tex]0^{\circ}C[/tex]
[tex]E_2=mc(\Delta T)\quad [c=\text{specific heat of water,}1\ cal./gm.^{\circ}C]\\E_2=6\times 1\times 100=600\ cal.[/tex]
Calories released during the conversion of water to the ice at [tex]0^{\circ}C[/tex]
[tex]E_3=mL_f\quad [L_f=\text{latent heat of fusion}]\\E_3=6\times 79.7=478.2\ cal.[/tex]
The total energy released is
[tex]E=E_1+E_2+E_3\\E=3198+600+478.2=4276.2\ cal.[/tex]