Olaf the snowman has an heat capacity of ice while the coffee can be
assumed to have the same heat capacity of water.
Reasons:
The heat lost by the coffee = The heat gained by Olaf...(1)
ΔQ = m₁·c₁·(T₃ - T₁) = m₂·c₂·(T₂ - T₃)
Mass of the snowman, m₁ = 38 kg
Initial temperature of the snowman, T₁ = -4.2°C
Volume of coffee, V = 15 mL = 0.000015 m³
Initial temperature of the coffee = 80.0 °C
Specific heat capacity of the snowman, c₁ = 2.097 J/(g·°C)
Specific heat capacity of the coffee, c₂ = 4.187 J/(g·°C)
Latent heat of fusion of the coffee into ice = 334 J/g
Density of the ice, ρ₂ = 997 kg/m³
Solution:
Mass of the coffee, m₂ = 0.000015 m³ × 997 kg/m³ = 0.014955 kg = 14.955 grams
Heat given to convert the coffee to ice = 334 J/g × 14.955 g = 4,994.97 J
Heat given to cool the coffee to 0°C = 14.955×4.187×80.0 = 5009.3268
Heat given to cool the coffee to 0°C = 5,009.3268 J
Heat given to reduce the temperature of the coffee turned to ice is
presented as follows;
Heat to cool frozen coffee = 14.955 × 2.108 × T₃ = 31.52514·T₃
Heat gained by Olaf = m₁·c₁·(T₃ - T₁) = 38,000× 2.097 × (T₃ - (-4.2))
Which from equation (1) gives
4,994.97 + 5009.3268 + 31.52514·T₃ = 38,000× 2.097 × (T₃ - (-4.2))
Solving with a graphing calculator, gives;
T₃ ≈ -4.08 °C
The final temperature of Olaf, T₃ ≈ -4.08 °C
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