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9. You have 0.5 kg of water at 20°C in a calorimeter. You add 100 g of ice at 0°C. After all the ice melts, the final temperature of the water is 5°C. Calculate the specific latent heat of fusion of ice. (Assume no heat loss to the surroundings and the specific heat capacity of water is 4184 J/kg°C)?
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9. You have 0.5 kg of water at 20°C in a calorimeter. You add 100 g of...
Understanding the Problem
You have 0.5 kg of water at 20°C and add 0.1 kg of ice at 0°C. The final temperature after the ice melts is 5°C. We need to determine the specific latent heat of fusion of ice.
Key Concepts
- Specific Heat Capacity of Water: 4184 J/kg°C
- Latent Heat of Fusion: Energy required to change ice at 0°C to water at 0°C without changing temperature.
Energy Calculations
1. Heat Lost by Water:
- Initial temperature (T1) = 20°C
- Final temperature (T2) = 5°C
- Mass of water = 0.5 kg
- Heat lost (Q_water) = mass x specific heat x temperature change
- Q_water = 0.5 kg x 4184 J/kg°C x (20°C - 5°C)
- Q_water = 0.5 x 4184 x 15 = 31380 J
2. Heat Gained by Ice:
- Mass of ice = 0.1 kg
- Heat gained (Q_ice) = mass x latent heat of fusion + mass x specific heat x temperature change
- Q_ice = (mass of ice x latent heat of fusion) + (mass of melted ice x specific heat x temperature change)
After melting, the ice warms to final temperature (5°C):
- Q_ice = m_ice * L_f + m_ice * c * (T2 - T_initial_ice)
- Q_ice = 0.1 kg * L_f + 0.1 kg * 4184 J/kg°C * (5°C - 0°C)
3. Equating Heat Lost and Gained:
- 31380 J = 0.1 kg * L_f + 0.1 kg * 4184 J/kg°C * 5°C
- 31380 J = 0.1 kg * L_f + 2092 J
Solving for Latent Heat of Fusion (L_f)
- Rearranging the equation:
- 31380 J - 2092 J = 0.1 kg * L_f
- 29288 J = 0.1 kg * L_f
- L_f = 292880 J/kg
Conclusion
The specific latent heat of fusion of ice is approximately 292880 J/kg.
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9. You have 0.5 kg of water at 20°C in a calorimeter. You add 100 g of ice at 0°C. After all the ice melts, the final temperature of the water is 5°C. Calculate the specific latent heat of fusion of ice. (Assume no heat loss to the surroundings and the specific heat capacity of water is 4184 J/kg°C)?
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9. You have 0.5 kg of water at 20°C in a calorimeter. You add 100 g of ice at 0°C. After all the ice melts, the final temperature of the water is 5°C. Calculate the specific latent heat of fusion of ice. (Assume no heat loss to the surroundings and the specific heat capacity of water is 4184 J/kg°C)? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about 9. You have 0.5 kg of water at 20°C in a calorimeter. You add 100 g of ice at 0°C. After all the ice melts, the final temperature of the water is 5°C. Calculate the specific latent heat of fusion of ice. (Assume no heat loss to the surroundings and the specific heat capacity of water is 4184 J/kg°C)? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 9. You have 0.5 kg of water at 20°C in a calorimeter. You add 100 g of ice at 0°C. After all the ice melts, the final temperature of the water is 5°C. Calculate the specific latent heat of fusion of ice. (Assume no heat loss to the surroundings and the specific heat capacity of water is 4184 J/kg°C)?.
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