temp. and heat?

Question

An ice-cube tray contains 0.356 kg of water at 18.0 C

How much heat must be removed from the water to cool it to 0 C and freeze it? Express your answer in joules.

Answer 1

Heat to be removed to cool water to 0° C
= (0.356) * 4180 * (18.0 - 0) = 26785.4 J

Heat to be removed to freeze water to ice at 0° C
= (0.356) * 334400 =
= 119046.4 J

Total heat to be removed
= 26785.4 + 119046.4
= 145832 J.

Answer 2

If I remember the Cp correctly, it takes 4180 joules to be removed to cool 1 kg of water 1 deg C. So to do this, q= W * Cp * (T), where W is 0.356 kg, Cp is 4180, and T is 18 degC.

To that you add the latent heat removed to freeze the water. If this is Qf in joules/kg, the heat removed will be Qf * 0.356 kg.

Answer 3

Q = m.s.dT + m.s.L where m is the mass, s the specitic heat and dT the temperature difference. Add the latent heat of solidification L of the quantity. Use the right units.

Answer 4

0.356 kg=356 gram
each 1 gram of water loses one calory to cool one degree of centigrade. so 356x18=6408 calory is the amount of calory which you should get from water to change it from 18 degree centigrade to zero centigrade water.
again for each one gram of zero degree water the amount of calory should be taken to change it to zero degree of ice is 85 calory. so 356x18x85=544680 is the amount of calory to change the zero degree water to zero degree ice.
the total calory is 544680+6408=551088
each calory equals 4.18 jules. so 551088x4.18=2303547.84 is the answer of your question.