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Heat - Thermos Flask

Grade 7ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Thermos flask, also known as a Vacuum flask, is a device designed to keep hot liquids hot and cold liquids cold for a long time by minimizing heat transfer through conduction, convection, and radiation.

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The flask consists of a double-walled glass or stainless steel vessel. The space between the two walls is evacuated to create a vacuum, which prevents heat loss or gain by conduction and convection as these processes require a material medium.

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To minimize heat transfer by radiation, the inner surface of the outer wall and the outer surface of the inner wall are silvered. The shiny surface reflects heat back into the flask (if the liquid is hot) or away from the flask (if the liquid is cold).

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The mouth of the flask is closed with a stopper made of an insulating material like cork or plastic to prevent heat loss by convection and evaporation, and to stop conduction through the top.

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The vessel is placed in a plastic or metal case with insulating supports (like felt or rubber pads) at the bottom to protect the glass and further reduce conduction of heat.

📐Formulae

Q=mcΔTQ = mcΔT

Heat Lost=Heat Gained\text{Heat Lost} = \text{Heat Gained}

Rate of Heat Transfer∝Temperature Difference(ΔT)\text{Rate of Heat Transfer} \propto \text{Temperature Difference} (\Delta T)

💡Examples

Problem 1:

Explain why the walls of a vacuum flask are silvered.

Solution:

The walls are silvered to minimize heat transfer by radiation.

Explanation:

Shiny surfaces are poor absorbers and poor emitters of heat, but excellent reflectors. The silvered surfaces reflect the thermal radiation back towards the source (e.g., reflecting heat back into a hot liquid), thereby significantly reducing heat loss or gain through the vacuum via electromagnetic waves.

Problem 2:

How does the vacuum between the double walls of a thermos flask help in maintaining the temperature of the liquid inside?

Solution:

The vacuum eliminates the medium required for conduction and convection.

Explanation:

Conduction and convection both require molecules to transfer heat energy. By removing the air between the walls (creating a vacuum), heat cannot travel from the inner wall to the outer wall (or vice versa) through these two processes, effectively insulating the contents.

Problem 3:

Calculate the heat energy required to raise the temperature of 500 g500\text{ g} of water inside a flask from 20∘C20^\circ\text{C} to 50∘C50^\circ\text{C}. (Specific heat capacity of water c=4.2 J/g∘Cc = 4.2\text{ J/g}^\circ\text{C})

Solution:

Q=63,000 JQ = 63,000\text{ J}

Explanation:

Using the formula Q=mcΔTQ = mc\Delta T: Given m=500 gm = 500\text{ g}, c=4.2 J/g∘Cc = 4.2\text{ J/g}^\circ\text{C}, and ΔT=50∘C−20∘C=30∘C\Delta T = 50^\circ\text{C} - 20^\circ\text{C} = 30^\circ\text{C}. Substituting the values: Q=500×4.2×30=63,000 JoulesQ = 500 \times 4.2 \times 30 = 63,000\text{ Joules}.