Answer:
The required mass of ice is 12.5 kg.
Explanation:
Mass of hot tea, [tex]m_{t}[/tex] = 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, [tex]c_{t}[/tex] = Specific heat capacity of water = 4190 J/kg/K
Heat lost tea = Heat gained by ice
[tex]m_{t}[/tex][tex]c_{t}[/tex]Δ[tex]T_{t}[/tex] = [tex]m_{i}[/tex]L + [tex]m_{i}[/tex][tex]c_{i}[/tex]Δ[tex]T_{t}[/tex]
[tex]m_{t}[/tex][tex]c_{t}[/tex]Δ[tex]T_{t}[/tex] = [tex]m_{i}[/tex] (L + [tex]c_{i}[/tex]Δ[tex]T_{t}[/tex])
1.8 x 4190 x (353 - 283) = [tex]m_{i}[/tex](334 + (4190 x (283 - 2730))
1.8 x 4190 x 70 = [tex]m_{i}[/tex](334 + (4190 x 10))
527940 = 42234[tex]m_{i}[/tex]
[tex]m_{i}[/tex] = [tex]\frac{527940}{42234}[/tex]
= 12.5004
[tex]m_{i}[/tex] = 12.5 kg
The required mass of ice is 12.5 kg.
The required amount of ice to bring the mixture to 10°C is 12.5 kg.
Given data:
The mass of hot tea is, m' = 1.8 kg.
The initial temperature of hot tea is, T = 80°C = 353 K.
The initial temperature of ice is, T' = 0°C = 273 K.
The final temperature of mixture is, T'' = 10°C = 283 K.
The heat of fusion of ice is, h = 334 kJ/kg.
The specific heat of water is, c = 4190 J/kg·K.
Apply the condition of thermal equilibrium which says that the heat released by the hot tea will be gained by the ice.
Heat lost by tea = Heat gained by ice
[tex]m'c(T-T'') = mh +mc(T''- T')[/tex]
Here, m is the mass of ice.
Solving as,
[tex]1.8 \times 4190 \times (353-283) = (m \times 334) +(m \times 4190(283- 273))\\\\527940 = 42234m\\\\m = 12.5 \;\rm kg[/tex]
Thus, we can conclude that the required amount of ice to bring the mixture to 10°C is 12.5 kg.
Learn more about the thermal equilibrium here:
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