Question

A person makes ice tea by adding ice to 1.8 kg of hot tea, initially at 80°C. How many kilograms of ice, initially at 0.00°C, are required to bring the mixture to 10°C? The heat of fusion of ice is 334 kJ/kg, and we can assume that tea has essentially the same thermal properties as water, so its specific heat is 4190 J/(kg·K).

Answer 1

Answer:

The required mass of ice is 12.5 kg.

Explanation:

Mass of hot tea, $m_{t}$ = 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, $c_{t}$ = Specific heat capacity of water = 4190 J/kg/K

Heat lost tea = Heat gained by ice

$m_{t}$$c_{t}$Δ$T_{t}$ = $m_{i}$L +  $m_{i}$$c_{i}$Δ$T_{t}$

$m_{t}$$c_{t}$Δ$T_{t}$ = $m_{i}$ (L +  $c_{i}$Δ$T_{t}$)

1.8 x 4190 x (353 - 283) = $m_{i}$(334 + (4190 x (283 - 2730))

1.8 x 4190 x 70 = $m_{i}$(334 + (4190 x 10))

527940 = 42234$m_{i}$

$m_{i}$ = $\frac{527940}{42234}$

    = 12.5004

$m_{i}$ = 12.5 kg

The required mass of ice is 12.5 kg.