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Ch. 19 - Heat and the First Law of Thermodynamics
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoGiancoli Douglas 5th editionCh. 19 - Heat and the First Law of ThermodynamicsProblema 70
Capitolo 19, Problema 70

A ceramic teapot (e = 0.70) and a shiny metal one (e = 0.10) each hold 0.55 L of tea at 85°C. (a) Estimate the rate of heat loss from each, and (b) estimate the temperature drop after 30 min for each. Consider only radiation, and assume the surroundings are at 20°C.

Guida verificata passo dopo passo
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Step 1: Understand the problem. The problem involves heat loss due to radiation, which is governed by the Stefan-Boltzmann law: P = eσA(T⁴ - Tₛ⁴), where P is the power (rate of heat loss), e is the emissivity, σ is the Stefan-Boltzmann constant (5.67 × 10⁻⁸ W/m²·K⁴), A is the surface area, T is the temperature of the object in Kelvin, and Tₛ is the temperature of the surroundings in Kelvin.
Step 2: Convert the given temperatures to Kelvin. Use the formula T(K) = T(°C) + 273.15. For the tea, T = 85 + 273.15, and for the surroundings, Tₛ = 20 + 273.15.
Step 3: Estimate the surface area of the teapots. Assume the teapots are approximately spherical. The volume of tea is given as 0.55 L (or 0.55 × 10⁻³ m³). For a sphere, the volume is V = (4/3)πr³, and the surface area is A = 4πr². Solve for the radius r from the volume, then calculate the surface area A.
Step 4: Calculate the rate of heat loss (P) for each teapot using the Stefan-Boltzmann law. Substitute the emissivity (e = 0.70 for the ceramic teapot and e = 0.10 for the shiny metal teapot), the calculated surface area (A), the Stefan-Boltzmann constant (σ), and the temperatures (T and Tₛ) into the formula P = eσA(T⁴ - Tₛ⁴).
Step 5: Estimate the temperature drop after 30 minutes. Use the relationship Q = P × t, where Q is the heat lost, P is the power (rate of heat loss), and t is the time in seconds (30 minutes = 1800 seconds). Then, use the formula Q = mcΔT, where m is the mass of the tea (density of water is approximately 1000 kg/m³, so m = 0.55 kg), c is the specific heat capacity of water (c = 4186 J/kg·K), and ΔT is the temperature change. Solve for ΔT for each teapot.

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Concetti chiave

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Emissivity

Emissivity is a measure of an object's ability to emit thermal radiation compared to a perfect black body, which has an emissivity of 1. In this context, the ceramic teapot has a higher emissivity (0.70) than the shiny metal teapot (0.10), meaning it will radiate heat more effectively. This property significantly influences the rate of heat loss through radiation.
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13:08
Radiation

Stefan-Boltzmann Law

The Stefan-Boltzmann Law states that the power radiated by a black body is proportional to the fourth power of its absolute temperature. For real objects, this law is modified by the emissivity factor. In this problem, it helps calculate the rate of heat loss from each teapot by considering their temperatures and emissivities, allowing us to estimate how much heat they lose to the surroundings.
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Gauss' Law

Heat Transfer

Heat transfer refers to the movement of thermal energy from one object to another due to a temperature difference. In this scenario, the teapots lose heat to the surrounding environment at a temperature of 20°C. Understanding the mechanisms of heat transfer, particularly radiation in this case, is essential for estimating the temperature drop of the tea over time.
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Overview of Heat Transfer
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