Mixing 100g water at 2°C with 50g ice at -4°C

[Raerin]

If 100g of water at 2 degrees Celsius is mixed with 50g of ice a -4 degrees Celsius, what is the final temperature of water? It also says that I need the heat of fusion of the ice to solve it, which I have found to be 334 kj/kg. I don't know what to do with the heat of fusion. Also, how do I convert 344 kj/kg to j/g? Is it still 334?

 

[Evgeny.Makarov]

how do I convert 344 kj/kg to j/g? Is it still 334?

Yes.

If 100g of water at 2 degrees Celsius is mixed with 50g of ice a -4 degrees Celsius, what is the final temperature of water? It also says that I need the heat of fusion of the ice to solve it, which I have found to be 334 kj/kg. I don't know what to do with the heat of fusion.

There are four options:

(1) all ice will melt
(2) some ice will melt
(3) some water will freeze
(4) all water will freeze

In cases (2) and (3) the resulting temperature of the mixture is 0 degrees Celsius.

Consider the following energies.

$Q_1$ heats 50g of ice from -4°C to 0°C.

$Q_2$ melts 50g of ice at 0°C.

$Q_3$ melts 100g of ice at 0°C, or is produced when 100g of water freezes.

$Q_4$ heats 100g of water from 0°C to 2°C, or is produces when 100g of water cools from 2°C to 0°C.

Then the options above take place under the following conditions.

(1) $Q_1+Q_2<Q_4$
(2) $Q_1<Q_4<Q_1+Q_2$
(3) $Q_4<Q_1<Q_4+Q_3$
(4) $Q_3+Q_4<Q_1$

$Q_2$ and $Q_3$ are found by multiplying heat of fusion by mass.