Answer:
The required mass of ice is 12.5 kg.
Explanation:
Mass of hot tea,
= 1.8 kg
Initial temperature of hot tea = 80°C = 353 K
Initial temperature of ice = 0.00°C = 273 K
Final temperature of mixture = 10°C = 283 K
Heat of fusion of ice, L = 334 kJ/kg
Specific heat capacity of tea,
= Specific heat capacity of water = 4190 J/kg/K
Heat lost tea = Heat gained by ice

Δ
=
L + 
Δ

Δ
=
(L +
Δ
)
1.8 x 4190 x (353 - 283) =
(334 + (4190 x (283 - 2730))
1.8 x 4190 x 70 =
(334 + (4190 x 10))
527940 = 42234
= 
= 12.5004
= 12.5 kg
The required mass of ice is 12.5 kg.