Answer:
Final Temperature of the drink = -15 °C
Explanation:
Heat lost by the water = heat gained by the cubes................. Equation 1
Heat lost by water = c₁m₁(T₁-T₃)..................... Equation 2
Heat gained by the cubes = Latent heat of fusion of the cubes + Heat capacity of the cubes.
⇒ Heat gained by the cubes = 2(lm₂) + 2{c₂m₂(T₂-T₃)}................... equation 3
Substituting equation 2 and 3 into equation 1
c₁m₁(T₁-T₃) = 2(lm₂) + 2{c₂m₂(T₃-T₂)}................................ Equation 4
Making T₃ The subject of formula in equation 4 above,
T₃ = (c₁m₁T₁ - 2lm₂ + 2c₂m₂T₂)/(c₁m₁ + 2c₂m₂)................... Equation 5
Where c₁ = Specific heat capacity of water, c₂ = specific heat capacity of ice, m₁ = mass of water, m₂ = mass of ice, l = specific latent heat of fusion of ice, T₁ = initial temperature of water, T₂ = initial temperature of ice, T₃ = final temperature of the drink.
Constant:<em> l = 336000 J/kg, c₁ = 4200 J/kg.K, c₂ = 2100 J/kg.K.</em>
<em>Given: m₁= 200 g = (200/100) kg = 0.2 kg, m₂ = 50 g = (50/1000) kg = 0.05 kg, T₁ = 25 °C = 298 K, T₂ = -15 °C = 258 K</em>
Substituting these values into equation 5,
T₃ = (4200×0.2×298 + 2×336000×0.05 + 2×2100×0.05×258)/(4200×0.2 + 2×0.05×2100)
T₃ = (250320 -33600 + 54180)/(840 + 210)
T₃ =270900/1050
T₃ = 258 K = - 15 °C
Therefore The Final Temperature of the drink = -15 °C
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