Answer:
The thermal energy gained by the water during the entire process is 90,372 J
Explanation:
Given;
mass of the ice cube, m = 30 g
latent heat of fusion of ice, L = 334 J/g
specific heat capacity of water, C = 4.184 J/g⁰C
latent heat of vaporization of water, h = 2260 J/g
The thermal energy gained by the water during the entire process is calculated as;
[tex]Q_t = Q_{fusion-ice} \ + \ Q_{boil-water} \ + \ Q_{steam-vapor}[/tex]
The heat of fusion of the ice at 0⁰C is calculated as;
[tex]Q_{fusion} = mL = 30\ g \times 334 J/g = 10020 \ J[/tex]
The heat of capacity of the water from 0⁰C to 100 ⁰C
[tex]Q_{boil-water} = mc\Delta \theta = mc(100-0) = 30\ g \times 4.184\ J/g^0C \times 100\ ^0C = 12,552 \ J[/tex]
The heat of vaporization of the steam at 100⁰C is calculated as;
[tex]Q_{steam} = mh= 30 g \times 2260 \ J/g= 67,800 \ J[/tex]
The thermal energy gained by the water during the entire procees;
Qt = 10,020 J + 12,552 J + 67,800 J
= 90,372 J
Therefore, the thermal energy gained by the water during the entire process is 90,372 J
