Heat Transfer Formula

Heat transfer is the fascinating phenomenon where kinetic energy (energy associated with the microscopic motion of atoms and molecules) moves between different objects or systems due to a temperature difference. Think of it like this:

  • Hotter objects have molecules that vibrate and move faster, holding more thermal energy.
  • Cooler objects have slower-moving molecules with less thermal energy.

Heat transfer works to balance this difference, trying to reach an equilibrium where both objects have the same temperature. This exchange can happen in three main ways:

  1. Conduction
  2. Convection
  3. Radiation

Heat Transfer Formula

The heat transfer formula is a fundamental equation used to quantify the amount of heat exchanged between two systems or objects. It is expressed as:

$$\begin{array}{l}Q= c\times m\times \Delta T\end{array}$$

Where:

  • represents the amount of heat transferred.
  • is the specific heat capacity of the material.
  • is the mass of the object.
  • is the change in temperature of the system.

Concept of Heat Transfer

Heat transfer is a fundamental concept in physics, engineering, and even everyday life. It explains how thermal energy (energy associated with the microscopic motion of atoms and molecules) moves between different objects or systems due to a temperature difference.

Here’s a breakdown of the key aspects:

The Driving Force:

  • All objects hold thermal energy, related to the average kinetic energy of their atoms and molecules.
  • Temperature reflects this average kinetic energy – higher temperatures indicate faster-moving molecules with more kinetic energy.
  • Heat transfer occurs because of this temperature difference. Hotter objects tend to lose thermal energy, while cooler objects tend to gain it, driven by the natural tendency to reach thermal equilibrium (equal temperatures).

Modes of Heat Transfer



Here are the three different methods of heat transfer:

1. Conduction

This occurs when heat is transferred directly between objects that are in physical contact. The transfer of heat is due to the collisions of microscopic particles (atoms and molecules) between the two objects. The hotter object has more energetic particles, and when they collide with the particles of the cooler object, they transfer some of their energy. This process continues until the two objects reach the same temperature.

Heat transferred by the process of conduction:$$\begin{array}{l}Q= \frac{kA\left ( T_{Hot}-T_{Cold} \right )_{t}}{d}\end{array}$$

Where;

  • Q = Heat transferred
  • σ = Stefan Boltzmann Constant
  • T(Hot)= Hot temperature
  • T(Cold) = Cold Temperature
  • A = Area of surface
  • d = Thickness of the material

2. Convection

This occurs when heat is transferred through a fluid (liquid or gas) that is moving. The heated fluid expands and becomes less dense, so it rises. Cooler fluid then sinks to take its place. This circulation of fluid creates a continuous flow of heat. Convection is responsible for heating the air in a room with a radiator, for example.

Heat transferred by the process of convection:$$\begin{array}{l}Q = H_{c}A\left ( T_{HOT}-T_{COLD} \right )\end{array}$$

Where;

  • Q = rate of heat transfer;
  • HC = Heat Transfer Coefficient
  • T(Hot) = Hot temperature
  • T(Cold) = Cold Temperature
  • A = Area of surface

3. Radiation

This occurs when heat is transferred in the form of electromagnetic waves. Unlike conduction and convection, radiation does not require any physical contact between the objects. All objects emit electromagnetic waves, and the hotter an object is, the more radiation it emits. The waves can travel through a vacuum, which is why we can feel the heat from the sun even though there is no air in space.

Heat transferred by the process of radiation:$$\begin{array}{l}Q= \sigma \left ( T_{4}^{Hot}-T_{4}^{Cold} \right )A\end{array}$$

Where,

  • Q = Heat transferred
  • σ = Stefan Boltzmann Constant
  • T(Hot)= Hot temperature
  • T(Cold) = Cold Temperature
  • A = Area of surface

Practice Problems on Heat Transfer Formula

Conduction:

1. A metal spoon initially at room temperature (20°C) is dipped into a cup of hot coffee at 80°C. The spoon is made of steel, which has a thermal conductivity of 45 W/mK. If the spoon has a length of 10 cm and a diameter of 5 mm, how much heat does it absorb in 1 minute?

Given:

  • Spoon material: Steel (k = 45 W/mK)
  • Spoon length: 10 cm = 0.1 m
  • Spoon diameter: 5 mm = 0.005 m (radius = 0.0025 m)
  • Coffee temperature: 80°C
  • Room temperature: 20°C
  • Time: 1 minute = 60 seconds

Solution:

  1. Calculate the area of the spoon’s cylindrical surface: A = 2πrl = 2π * 0.0025 * 0.1 = 0.0157 m²
  2. Calculate the temperature difference: ΔT = 80°C – 20°C = 60°C
  3. Set up the heat transfer equation for conduction: Q = kAΔT / L
  4. Solve for the heat absorbed: Q = 45 W/mK * 0.0157 m² * 60°C / 0.1 m = 41.03 J

2. A double-paned window consists of two glass panes each 5 mm thick, separated by an air gap of 1 cm. The inside temperature of the window is 20°C, and the outside temperature is -5°C. What is the rate of heat transfer through the window if the total area of each pane is 1 m²? Assume the thermal conductivity of air is 0.025 W/mK and the thermal conductivity of glass is 1 W/mK.

  • Glass thickness: 5 mm = 0.005 m
  • Air gap thickness: 1 cm = 0.01 m
  • Glass thermal conductivity: 1 W/mK
  • Air thermal conductivity: 0.025 W/mK
  • Inside temperature: 20°C
  • Outside temperature: -5°C
  • Window area: 1 m² (per pane)

Solution:

  1. Calculate the total thermal resistance of the window: Rtotal = R1 + R2
  2. Calculate the resistance of each glass pane: R1 = d / k = 0.005 m / 1 W/mK = 0.005 K/W
  3. Calculate the resistance of the air gap: R2 = d / k = 0.01 m / 0.025 W/mK = 0.4 K/W
  4. Add the resistances to find the total: Rtotal = 0.005 K/W + 0.4 K/W = 0.405 K/W
  5. Calculate the temperature difference: ΔT = 20°C – (-5°C) = 25°C
  6. Use the formula Q = ΔT / R to find the heat transfer rate: Q = 25°C / 0.405 K/W = 61.73 W

Convection:

  1. A cup of hot tea initially at 90°C is placed in a room with a temperature of 20°C. The convection heat transfer coefficient between the tea and the surrounding air is 10 W/m²K. If the surface area of the tea is 0.05 m², how long will it take for the tea to cool down to 50°C?

Given:

  • Initial tea temperature: 90°C
  • Room temperature: 20°C
  • Convection coefficient: 10 W/m²K
  • Tea surface area: 0.05 m²

Solution:

  1. Set up the Newton’s law of cooling equation: Q = hAΔT
  2. Solve for the time it takes to cool down to 50°C: t = Q / (hAΔT) = (m * c * ΔT) / (hAΔT)
    • Assuming specific heat of water (c) = 4186 J/kg°C and water mass (m) is irrelevant since Q cancels out.
  3. Plug in the values: t = (0.05 kg * 4186 J/kg°C * (90°C – 50°C)) / (10 W/m²K * 0.05 m²) = 727.2 seconds (12.12 minutes)

Therefore, it will take approximately 727.2 seconds (12.12 minutes) for the tea to cool down to 50°C.

Radiation:

  1. A person has a body temperature of 37°C and a surface area of 1.5 m². Assuming the person’s emissivity is 0.95 and the surrounding temperature is 20°C, what is the rate of heat loss from the person’s body due to radiation?

Given:

  • Body temperature: 37°C = 310 K
  • Surface area: 1.5 m²
  • Emissivity: 0.95
  • Surroundings temperature: 20°C = 293 K

Solution:

  1. Use the Stefan-Boltzmann law to calculate the radiated power per unit area:
P = εσT⁴

where:

  • P is the radiated power per unit area (W/m²)
  • ε is the emissivity (dimensionless)
  • σ is the Stefan-Boltzmann constant (5.67 × 10^-8 W/m²K⁴)
  • T is the absolute temperature (K)
  1. Plug in the given values:
P = 0.95 * 5.67 × 10^-8 W/m²K⁴ * (310 K)⁴ = 505.38 W/m²
  1. Calculate the total radiated power by multiplying the power per unit area by the surface area:
Q = P * A
Q = 505.38 W/m² * 1.5 m² = 758.07 W

Therefore, the rate of heat loss from the person’s body due to radiation is approximately 758.07 watts.

Note: This calculation assumes the person is not wearing any clothing and is in a completely enclosed environment. The actual heat loss can be affected by various factors like clothing, air movement, and humidity.

Heat Transfer Practice Problems:

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HEAT TRANSFER QUIZ

Test your understanding and challenge yourself with these questions.

1 / 8

Which of the following is not a method of heat transfer?

2 / 8

True or False: Vacuum flasks prevent heat loss or gain by utilizing all three methods of heat transfer (conduction, convection, and radiation).

3 / 8

In which material does conduction occur most efficiently?

4 / 8

In terms of heat transfer, why are metals generally cooler to the touch than wood, if both are at the same temperature?

5 / 8

What is the primary method of heat transfer in liquids and gases?

Convection is the primary method of heat transfer in fluids (liquids and gases), where warmer portions of the fluid rise and cooler portions sink, creating a circulating flow.

6 / 8

Which factor does not affect the rate of heat transfer by conduction?

7 / 8

Which type of radiation is primarily responsible for heat transfer from the Sun to the Earth?

8 / 8

When boiling water, the bubbles you see are due to which type of heat transfer?

Your score is

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Related Physics Formulas

FAQs (Frequently Asked Questions)

1. What is heat transfer formula?

The heat transfer formula quantifies the amount of heat transferred between two systems or objects. It is expressed as , where

  •  is the heat transferred,
  •  is the specific heat capacity of the material,
  •  is the mass of the object, and
  •  is the change in temperature of the system.

2. What is the Q equation for heat transfer?

The  equation for heat transfer is , where

  •  represents the amount of heat transferred,
  •  is the specific heat capacity of the material,
  •  is the mass of the object, and
  •  is the change in temperature of the system.

3. What is the formula for heat transfer in gate?

4. What is the SI unit of heat?

5. What are the 4 methods of heat transfer?