[GOPgabe]

I don't any Thermochemistry so forgive my ignorance on this question.
An ice cube at 0.00 °C is placed in 200. g of distilled water at 25.00 °C. The final temperature after the ice is completely melted is 5.00 °C.
What is the mass of the ice cube? (ΔHfus = 340. J.g-1, Cp = 4.18 J.g-1.°C-1)

Answer=46.3 grams.

I tried heat lost = heat gained, but I'm not sure if this method is correct as I keep coming out with 49 g, which is not the correct answer. Any aid would be much appreciated.

[opti384]

The heat lost from the distilled water should be 200g x 4.18 J.g-1.°C-1 x (25-5)°C
Then the heat gained by the ice cube is : the mass of ice cube x 340. J. g-1

Therefore you get             200g x 4.18 J.g-1.°C-1 x (25-5)°C = the mass of ice cube x 340. J. g-1
from the above equation you get 49.2 g for the mass of the ice cube.

[GOPgabe]

The heat lost from the distilled water should be 200g x 4.18 J.g-1.°C-1 x (25-5)°C
Then the heat gained by the ice cube is : the mass of ice cube x 340. J. g-1

Therefore you get             200g x 4.18 J.g-1.°C-1 x (25-5)°C = the mass of ice cube x 340. J. g-1
from the above equation you get 49.2 g for the mass of the ice cube.

I got that as well. The question came from one of the Chemistry Olympiad Local Selection tests. In this scenario maybe the answer key is wrong? Thanks very much for your response.

[A Tree]

The heat lost from the distilled water should be 200g x 4.18 J.g-1.°C-1 x (25-5)°C
Then the heat gained by the ice cube is : the mass of ice cube x 340. J. g-1

Therefore you get             200g x 4.18 J.g-1.°C-1 x (25-5)°C = the mass of ice cube x 340. J. g-1
from the above equation you get 49.2 g for the mass of the ice cube.

In fact, the heat gained by the ice cube includes two parts: 1) ice(0°C) -->water(0°C), and 2) water(0°C) -->water(5°C). You missed the latter part.

[opti384]

In fact, the heat gained by the ice cube includes two parts: 1) ice(0°C) -->water(0°C), and 2) water(0°C) -->water(5°C). You missed the latter part.
[/quote]

Yes I made a big mistake. It should be 200g x 4.18 J.g-1.°C-1 x (25-5)°C = the mass of ice cube x 340. J. g-1 + 4.18 x the mass of ice x 5°C and you'll get the answer 46.3 g. Thanks for the correction.

[GOPgabe]

Thanks for your help guys.