A spherical pot contains 0.750.75 L of hot coffee (essentially water) at an initial temperature of 9595°C. The pot has an emissivity of 0.600.60, and the surroundings are at 20.0 20.0°C. Calculate the coffee's rate of heat loss by radiation.

Ch 17: Temperature and Heat

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 17: Temperature and Heat

Problem 67

Chapter 17, Problem 67

A spherical pot contains $0.75$ L of hot coffee (essentially water) at an initial temperature of $95$ °C. The pot has an emissivity of $0.60$ , and the surroundings are at $20.0$ °C. Calculate the coffee's rate of heat loss by radiation.

Verified step by step guidance

First, understand that the rate of heat loss by radiation can be calculated using the Stefan-Boltzmann Law, which is given by the formula:

P = e σ A (T^{4} - T^{4})

, where

P

is the power (rate of heat loss),

e

is the emissivity,

σ

is the Stefan-Boltzmann constant (

5.67 \times 10 ⁻⁸ W / m^{2} K^{4}

A

is the surface area, and

T

are the temperatures in Kelvin.

Convert the temperatures from Celsius to Kelvin. The initial temperature of the coffee is 95°C, which is

95 + 273.15 = 368.15

K. The surrounding temperature is 20°C, which is

20 + 273.15 = 293.15

Calculate the surface area of the spherical pot. The volume of the pot is given as 0.75 L, which is equivalent to 0.75 × 10⁻³ m³. Use the formula for the volume of a sphere:

V = \frac{4}{3} π r^{3}

. Solve for the radius

r

, and then use the formula for the surface area of a sphere:

A = 4 π r^{2}

Substitute the values into the Stefan-Boltzmann Law formula. Use the emissivity

0.60

, the Stefan-Boltzmann constant

5.67 \times 10 ⁻⁸

, the calculated surface area, and the temperatures in Kelvin.

Calculate the difference in the fourth powers of the temperatures:

{368.15}^{4} - {293.15}^{4}

. Multiply this result by the emissivity, Stefan-Boltzmann constant, and surface area to find the rate of heat loss by radiation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law describes the power radiated from a black body in terms of its temperature. It states that the total energy radiated per unit surface area is proportional to the fourth power of the black body's temperature. For non-black bodies, emissivity is included to account for the material's efficiency in emitting radiation.

Emissivity

Emissivity is a measure of how effectively a surface emits thermal radiation compared to a perfect black body. It ranges from 0 to 1, where 1 represents a perfect emitter. In this problem, the pot's emissivity of 0.60 indicates it emits 60% of the radiation a black body would at the same temperature.

Temperature Difference

The rate of heat loss by radiation depends on the temperature difference between the object and its surroundings. The greater the difference, the higher the rate of heat transfer. Here, the coffee's initial temperature is 95°C, and the surroundings are at 20°C, creating a significant temperature gradient driving the heat loss.

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