When 200 grams of ice is added to a bucket containing 1.00 litre of hot water, what’s the final temperature of the water?

To answer the question, we’re going to need to make some assumptions. We’ll take 1.000 litre of pure water at 80.00°C and add 200.0 g of ice (at -10.00°C) to it. What’s the final temperature of the water?

## Part 1: Heat transfer method

The following equation can calculate the temperature at thermal equilibrium of any number of objects in thermal contact.

I love this equation because it’s several lines of maths shorter than the version taught in school. With this equation, you don’t even need to convert the temperatures into kelvin. Celsius works just fine.

Let’s set up the equation so that the addition series contains the variables in the question.

Now, let’s substitute the gives values into the equation. The specific heat capacity of water is 4200 J kg^{-1} K^{-1}, and that of ice is 2100 J kg^{-1} K^{-1}.

Great! Adding 200.0 g of ice to 1.000 L of water decreases the temperature from 80.00°C to 71.80°C.

But we’ve forgotten something. The ice will melt as soon as it hits the hot water. Since melting is an endothermic process, heat energy from the water will actually be absorbed, thus reducing the final temperature even further.

## Part 2: Let’s take into account the fact that the ice melts!

Remember our formula from part 1.

The amount of energy required to melt ice can be calculated using the latent heat equation:

Removing that amount of heat energy from the system results in the following equation:

Great! Now, we’ve calculated that the final temperature of the water would be 57.36°C after the addition of the ice. That’s equal to 330.5 kelvin, which will be useful later.

However, we’ve forgotten to take something else into account: how much heat will be lost as radiation from the surface of the bucket?

## Part 3: What’s the rate of heat loss from the bucket by radiation?

The rate of heat lost by radiation can be calculated by using the Stefan-Boltzmann equation, below.

P is the rate at which heat energy is radiated from the surface of the bucket in watts. Emissivity, e, of water is 0.95, and the surface area, A, should be around 0.0707 m^{2} for a one-litre bucket. Calculation of A is shown below. Assuming that the radius of the surface of the bucket is 6cm:

Plugging that value into the equation, we can find P. We’ll assume that the experiment is being conducted at room temperature and the temperature of the surroundings is 20.00°C (29.03 K).

This means that 2.928 joules of energy are emitted from the surface of the bucket every second. Ten minutes later, the bucket would have lost 1756.8 joules of energy due to radiation from the surface. But what about emission of radiation from the sides of the bucket?

Let’s say that our bucket is made from highly polished aluminium (which has emissivity 0.035) and it holds exactly 1.2 litres of water. We need to calculate the dimensions of the bucket.

Assuming it has straight sides (i.e. it’s a cylinder), the bucket had volume equal to the following formula:

The surface area of our bucket (excluding the open surface at the top) is:

The rate of energy radiation from the sides would therefore be:

It’s interesting to note how very little radiation is emitted from the shiny aluminium bucket, while lots more radiation is emitted from the surface of the water. This is because relatively ‘dark’ water has a much higher emissivity than shiny aluminium. Total emission from the bucket is therefore:

After ten minutes, the bucket would have lost the following amount of energy:

Let’s factor this amount of energy loss into our final temperature equation.

Not much energy is lost via radiation! Finally, let’s find the peak wavelength of the radiation emitted by the object using Wien’s law.

## Part 4: What’s the wavelength of the radiation being emitted by the bucket?

Here’s Wien’s law from Unit 1 Physics…

The radiation emitted from the resulting bucket of water lies firmly in the infra-red part of the electromagnetic spectrum. The bucket would be clearly visible on an infra-red camera!

Next week, we’ll begin a new a Chemistry-themed project called Periodic Table Smoothie. More next week.

Nice!

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Once the ice has reached 0C and gone through its phase change it becomes water. Therefore the specific heat capacity also changes to that of water. This example doesn’t account for this.

Qice = -Qwater

Qice = m(i)c(i)(0–10) + m(i)Lf(i) + m(i)c(w)(Tfinal-0)

Qwater = m(w)c(w)(Tfinal-80)

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