0.5kg of water at 10ºC is completely converted to ice at 0ºC by extracting (88000) of heat from it. If the…

Analyze the question: 0.5 kg of water at 10 °C is cooled to 0 °C and then frozen to ice at 0 °C. The total heat removed is 88,000 J (the unit is missing in the question, so I take it as joules). We need the specific latent heat of fusion of ice.

Concept: Heat removed has two parts:

1. Cooling the water from 10 °C to 0 °C: Q₁ = mcΔθ
2. Freezing the water at 0 °C: Q₂ = mL

Total: Q = mcΔθ + mL

Solution:

Q₁ = 0.5 × 4200 × 10 = 21,000 J

Q₂ = 88,000 − 21,000 = 67,000 J

L = Q₂ / m = 67,000 / 0.5 = 134,000 J/kg = 134 kJ/kg

Check: The real value for ice is about 334 kJ/kg, so the question's figure is not realistic.

Answer: C. 134.0 kJ/kg

Tip: In calorimetry questions, always split the process into stages: first the temperature change (mcΔθ), then the change of state (mL). Add the stages to get the total heat, then subtract to find the missing one. Also watch the units: 134,000 J/kg must be converted to kJ/kg to match the options.