An ice-cube tray of negligible mass contains 2.5 kg of water at 25°C. How much...

m = 0.29 kg

initial temperature = 14 deg C

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

heat to freeze it completely = mass * latent heat of fusion = $0.29 \times 334000 = 96860 J$

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