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Heat Transfer - Everyday Applications of Heat Transfer

Grade 8IB

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

🔑Concepts

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Heat transfer occurs from a region of higher temperature to a region of lower temperature through three main processes: Conduction, Convection, and Radiation.

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Conduction in Everyday Life: Cooking utensils are made of metals like copper or aluminum (good conductors) to transfer heat quickly to food, while handles are made of plastic or wood (insulators) for safety. Woollen clothes keep us warm by trapping air, which is a poor conductor of heat, preventing body heat from escaping (kair≈0.024 W/mKk_{air} \approx 0.024 \text{ W/mK}).

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Convection in Everyday Life: Convection currents explain Sea Breezes (daytime, air moves from sea to land) and Land Breezes (nighttime, air moves from land to sea). Heaters are placed on the floor because warm air is less dense and rises, whereas air conditioners are placed high because cool air is denser and sinks.

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Radiation in Everyday Life: Thermal radiation does not require a medium. Dark-colored clothes absorb more radiant heat than light-colored ones, which is why white clothes are preferred in summer. Solar water heaters use black pipes to maximize heat absorption.

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The Vacuum Flask (Thermos): Designed to minimize all three forms of heat transfer. The vacuum between the double glass walls prevents conduction and convection. The silvered inner surfaces reflect heat back, minimizing radiation. The plastic/cork stopper prevents heat loss by evaporation and convection.

📐Formulae

Q=m⋅c⋅ΔTQ = m \cdot c \cdot \Delta T

ΔT=Tfinal−Tinitial\Delta T = T_{final} - T_{initial}

P=QtP = \frac{Q}{t}

Efficiency=(Useful Energy OutputTotal Energy Input)×100%Efficiency = \left( \frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \right) \times 100\%

💡Examples

Problem 1:

A solar water heater absorbs 84000 J84000 \text{ J} of energy to heat 2 kg2 \text{ kg} of water. If the initial temperature is 25∘C25^{\circ}C, calculate the final temperature of the water. (Specific heat capacity of water c=4200 J/kg∘Cc = 4200 \text{ J/kg}^{\circ}C)

Solution:

Using the formula Q=m⋅c⋅ΔTQ = m \cdot c \cdot \Delta T, we can rearrange it to find ΔT\Delta T: ΔT=Qm⋅c\Delta T = \frac{Q}{m \cdot c} ΔT=840002⋅4200\Delta T = \frac{84000}{2 \cdot 4200} ΔT=840008400=10∘C\Delta T = \frac{84000}{8400} = 10^{\circ}C The final temperature is: Tfinal=Tinitial+ΔTT_{final} = T_{initial} + \Delta T Tfinal=25+10=35∘CT_{final} = 25 + 10 = 35^{\circ}C

Explanation:

The energy absorbed by the solar collector (radiation) is transferred to the water, increasing its internal energy and temperature.

Problem 2:

Compare the heat loss through a single-pane glass window versus a double-pane window with a vacuum layer.

Solution:

In a single-pane window, heat is lost via conduction through the glass: Qloss∝A⋅ΔTdQ_{loss} \propto \frac{A \cdot \Delta T}{d}. In a double-pane window, the vacuum layer eliminates conduction and convection between the panes because there are no particles to transfer kinetic energy. Only radiation can pass through, which is significantly slower.

Explanation:

This is an application of thermal insulation in green building design to maintain room temperature and reduce energy costs.

Problem 3:

Calculate the total energy required to heat an aluminum frying pan of mass 0.8 kg0.8 \text{ kg} from 30∘C30^{\circ}C to 180∘C180^{\circ}C. (cAl=900 J/kg∘Cc_{Al} = 900 \text{ J/kg}^{\circ}C)

Solution:

Step 1: Calculate the temperature change: 180−30150\begin{array}{r} 180 \\ -30 \\ \hline 150 \end{array} So, ΔT=150∘C\Delta T = 150^{\circ}C. Step 2: Use the heat formula: Q=0.8×900×150Q = 0.8 \times 900 \times 150 Q=720×150Q = 720 \times 150 Q=108000 JQ = 108000 \text{ J}

Explanation:

The high thermal conductivity and specific heat capacity of the metal allow it to store and transfer enough energy to cook food efficiently.