An ice-cube tray of negligible mass contains 2.5 kg of water at 25°C. How much heat must be removed to cool the water to 0°C and freeze it?

Clarence Juablar

Clarence Juablar

Answered question

2022-07-25

An ice-cube tray of negligible mass contains 2.5 kg of water at 25°C. How much heat must be removed to cool the water to 0°C and freeze it?

Answer & Explanation

Jeffrey Jordon

Jeffrey Jordon

Expert2022-11-07Added 2605 answers

m = 0.29 kg

initial temperature = 14 deg C

heat to bring the water to 0 deg C = mass*specific heat*change of temperature = 0.29 × 4184 × 14 = 16987.04 J

heat to freeze it completely = mass * latent heat of fusion = 0.29 × 334000 = 96860 J

 

a)So total heat removed injoules = 96860 + 16987.04 = 113847.04 J

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