Physics Problem that made me mad and it is for hw tmrw...?

Question

A jar of tea is placed in sunlight until it reaches an equilibrium temperature of 32 degrees celcius. In an attempt to cool the liquid, which has a mass of 180g , 112g of ice at 0 degrees celcius is added. At the time at which the temperature of the tea is 15 degrees celcius, determine the mass of the remaining ice in the jar. Assume the specific heat capacity of the tea to be that of pure liquid water.

Answer 1

Initially the liquid was at a temperature of 32 degree Celsius. If it comes to 15 degree Celsius then it's temperature has come down by 17 degrees of the same unit. If it's specific heat is the same as that of pure liquid water (which is 1 calorie per gram), then it has lost,

(180 gram) x (1 calorie\gram\Celsius) x 17 = 3060 calories.

Assuming that all the heat given out by the liquid is absorbed by the ice then after taking 180 calories let the mass of ice, that has now become liquid water, be m. The latent heat of ice is 80 calories\gram. Therefore,

m x (80 calories\gram) = 3060 calories
Hence,
m = (3060\80)gram
m = 38.25 gram.

The mass of ice that remains is

(112 - 38.25)gram

= 73.75 gram.

Which is the required quantity.

Answer 2

The tea will 'cool' at a rate of 1 calorie per gram per degree C.
Just looking at the original tea, how much energy has it lost to get to the new temperature.

Then that energy must have been absorbed by the ice (k calories per gram to melt + 1 cal. per gram per dec. C to rise to temperature). How many grams at k+15 cal. must have melted then risen to 15 C?

This, of course assumes that the rest of the ice is still frozen at 0 and that no energy was lost to evaporation.

For more fun, you could be able to find out how many calories per gram the jar is absorbing from the Sun (it is a formula that involves the 4th power of the equilibrium temperature in degrees K -- if that is not familiar to you, then I suppose that you are to ignore the effect of the sunlight on melting ice).

In reality, some of the melt water will not have reached 15 yet and some of the ice will be already in its melting process (some energy will have started to weaken the bonds). But ignore that (unless you are already at that stage of the course).

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Summary:

How many calories lost by the tea going from 32 to 15?
180 g * (32-15)C* [1 cal. / (g*C)] = A

This has to be equal to the energy needed to 'melt and raise to 15 C' a quantity (x grams) of ice. The remainder (112 - x) is still frozen.

k=80

x = A calories / (80 + 15) cal. per gram

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Once all the ice has melted, add a few drops of lemon juice and (optional) a bit of sugar. Enjoy the tea.

Answer 3

To bring the water from 32°C to 15°C it has to chamge by
17°C.
The specific heat of the tea (water) is 1cal/g/°C.
17 x 1 x 180 = 3060cal

This heat is removed from the ice.
The latent heat of melting ice is 80cal/g.

112 x 80 = 8960cal.
8960 - 3060 = 5900cal.
5900 ÷ 80 = 73.75 grams of ice remaining.