To analyze the cooling of the loaf of bread, we need to consider both convection and radiation heat transfer mechanisms. The loaf is initially at a temperature of 120 °C, and it will lose heat to the surrounding air at 20 °C. The convection coefficient is given as 10 W/m²K, and we also have the emissivity and thermal conductivity of the bread. Let's break down the problem step by step.

Understanding Heat Transfer Mechanisms

Heat transfer occurs through three primary mechanisms: conduction, convection, and radiation. In this scenario, we will focus on convection and radiation since the loaf is cooling in air.

Convection Heat Transfer

Convection is the process of heat transfer between a solid surface and a fluid (in this case, air) in motion. The rate of heat transfer by convection can be calculated using Newton's Law of Cooling:

Q_conv = h * A * (T_surface - T_air)

• Q_conv: Heat transfer rate by convection (W)
• h: Convection heat transfer coefficient (10 W/m²K)
• A: Surface area of the loaf (m²)
• T_surface: Surface temperature of the loaf (120 °C)
• T_air: Ambient air temperature (20 °C)

Calculating the Surface Area

To proceed, we need to determine the surface area of the loaf. Assuming the loaf is a rectangular prism, we can use the formula for the surface area of a rectangular box:

A = 2(lw + lh + wh)

Where:

• l: Length of the loaf
• w: Width of the loaf
• h: Height of the loaf

For example, if we assume the loaf dimensions are 0.3 m (length) x 0.1 m (width) x 0.1 m (height), the surface area would be:

A = 2(0.3*0.1 + 0.3*0.1 + 0.1*0.1) = 2(0.03 + 0.03 + 0.01) = 2(0.07) = 0.14 m²

Calculating Convection Heat Loss

Now, substituting the values into the convection formula:

Q_conv = 10 W/m²K * 0.14 m² * (120 °C - 20 °C)

Q_conv = 10 * 0.14 * 100 = 140 W

Radiation Heat Transfer

Next, we consider the heat loss due to radiation. The Stefan-Boltzmann Law gives us the rate of heat transfer by radiation:

Q_rad = ε * σ * A * (T_surface^4 - T_air^4)

• Q_rad: Heat transfer rate by radiation (W)
• ε: Emissivity of the bread (0.76)
• σ: Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/m²K^4)
• T_surface: Surface temperature in Kelvin (393 K)
• T_air: Ambient air temperature in Kelvin (293 K)

Converting the temperatures to Kelvin:

T_surface = 120 + 273 = 393 K

T_air = 20 + 273 = 293 K

Now substituting the values into the radiation formula:

Q_rad = 0.76 * (5.67 x 10^-8) * 0.14 * (393^4 - 293^4)

Calculating the fourth powers:

393^4 ≈ 2.426 x 10^9

293^4 ≈ 7.703 x 10^8

Now, substituting these values:

Q_rad = 0.76 * (5.67 x 10^-8) * 0.14 * (2.426 x 10^9 - 7.703 x 10^8)

Q_rad ≈ 0.76 * (5.67 x 10^-8) * 0.14 * (1.655 x 10^9)

Q_rad ≈ 0.76 * (5.67 x 10^-8) * 0.14 * 1.655 x 10^9 ≈ 0.76 * 0.0000000567 * 0.14 * 1655000000 ≈ 0.76 * 0.0000000567 * 231700000 ≈ 0.76 * 0.0131 ≈ 0.0099 W

Combining Heat Loss Mechanisms

Finally, to find the total heat loss from the loaf, we can sum the convection and radiation heat losses:

Total Heat Loss = Q_conv + Q_rad

Total Heat Loss = 140 W + 0.0099 W ≈ 140.0099 W

This total gives us a comprehensive view of how the loaf of bread cools down in the surrounding air. Understanding these principles not only helps in this specific scenario but also provides insights into broader applications in thermodynamics and heat transfer.