The 3 states of matter – solid, liquid and gas.
In general, when a body is heated, it expands and volume increases. The mass remains the same. Since density = mass/volume, its density decreases (less dense). For instance, warm air rises as it is less dense. In terms of kinetic theory, the particles will increase in kinetic energy. The average spacing between the particles increases (assuming not in a closed container).
Likewise, when a body is cooled, the opposite occurs. The body contracts and volume decreases. It becomes denser.
Due to the differences in particles arrangement of solid, liquid and gas, each expands by different amount when heated and vice versa. Which expands the most when heated and contracts the most when cooled?
The following demonstration of the ‘Pee Boy’ is a good video to show the concepts.
The tiny hole at the penis is too small for any water to enter on its own. So using thermal transfer in the different states, the following steps are taken:
Water boils at 100°C. In the view below, the water is initially at around 60°C.
How is this possible?
At sea level, where the pressure is atmospheric pressure (approx 105 000 Pa), the boiling point of water is 100°C.
In the video as air is sucked from the sealed container, the pressure in the container decreases. This lowers the boiling point of the water to around 60°C, hence water will boil and bubbles are formed.
It is similar to boiling water at high altitude, like on top of Mount Everest (about 8 km high). Water will boil around 70°C to to lower atmospheric pressure at high altitude.
This structure is located at Marina Bay, beside the Marina Bay Sands. It sprays out mist into the air. What is the main purpose of this in terms of thermal physics?
You might have come across similar features like this in other places like coffee shop, al fresco Starbucks etc.
Mist contains tiny droplets of water. When it is sprayed out to the surrounding, it will evaporate very fast.
The higher rate of evaporation is due to the tiny droplets of water having a relatively higher expose surface area compared to a body of water of the same mass.
As the droplets of water evaporate, thermal energy is taken from the surrounding air to change the water to a gaseous state. Hence temperature of the surrounding air drops, giving the cooler and conducive environment in a hot tropical weather in Singapore.
Four bars, all exactly the same size, are each placed with one end in boiling water.The times taken for the temperature of the other end to increase by 2 oC are measured.
|Material of bar||Time for 2oC rise / s|
To make a large metal tank with the least heat loss, which materials should be used for the tank and its insulation?
Solutions: Option B
This question can be a bit tricky. It depends on which approach you view the question. Simply using conduction will be a much easier way to get the answer. The hint to use conduction is from the first paragraph. Time taken for temp of the other end to increase by 2 oC is about conduction.
To build such a tank, metal (aluminium or copper) has to be used and insulation on the external wall (cork or styrofoam).
To contain boiling water with least heat loss, both tank and insulation have to be good insulator (poor conductor) to reduce heat lost to surrounding. Hence tank should be aluminium and insulation should be styrofoam (option B)
On the other hand, if you approach the question in term of heat capacity, it will be a bit tedious and you do not have the values.Firstly, if aluminium takes a longer time to rise by 2 oC, it has a higher specific heat capacity. But you cannot merely look at specific heat capacity. You have to look at heat capacity as the mass of the container is important.
Copper: Density = 8940 kg/m3 and specific heat capacity = 400 J/kg.K
Aluminium: Density = 2700 kg/m3 specific heat capacity = 900J/kg.K
Assumptions: Volume of copper and aluminium are the same (same shape of container) say 0.02 m3, Initial temperature of metal is 30oC
Considering the amount of thermal energy gained from water as metal temperature reaches 80oC
Heat lost by water = heat gain by metal
Mass of copper = 8940 x 0.02 = 178.8 kg
Heat lost be water = heat gained by copper = mcθ = 178.8 x 400 x (80-30) = 3576000 J
Mass of Aluminium = 2700 x 0.02 = 54 kg
Heat lost be water = heat gained by aluminium = mcθ = 54 x 900 x (80-30) = 2430000J
From the calculation, it is obvious that the aluminium gains lesser thermal energy from the water, hence water will remain warmer compared to when using copper. Hence using this approach, it is still Option B. But important to know that you have to consider heat capacity [C], not specific heat capacity (c). [ C = mc]