Question

How many calories are released when 6 grams of 100°C steam turns to 0°C ice?

Answer 1

Answer: 4276.2 calories

Step-by-step explanation:

Given

mass of steam is 6 gm at
`100^(\circ)C`

Conversion of steam to ice involves

• steam to water at
  `100^(\circ)C`

• water at
  `100^(\circ)C` to water at
  `0^(\circ)C`

• water to the ice at
  `0^(\circ)C`

Calories released during the conversion of steam to water at
`100^(\circ)C`

`E_1=mL_v\quad [L_v=\text{latent heat of vaporisation}]\\E_1=6* 533=3198\ cal.`

Calories released during the conversion of water at
`100^(\circ)C` to water at
`0^(\circ)C`

`E_2=mc(\Delta T)\quad [c=\text{specific heat of water,}1\ cal./gm.^(\circ)C]\\E_2=6* 1* 100=600\ cal.`

Calories released during the conversion of water to the ice at
`0^(\circ)C`

`E_3=mL_f\quad [L_f=\text{latent heat of fusion}]\\E_3=6* 79.7=478.2\ cal.`

The total energy released is

`E=E_1+E_2+E_3\\E=3198+600+478.2=4276.2\ cal.`