   Chapter 11, Problem 68AP

Chapter
Section
Textbook Problem

A wood stove is used to heat a single room. The stove is cylindrical in shape, with a diameter of 40.0 cm and a length of 50.0 cm, and operates at a temperature of 400.°F. (a) If the temperature of the room is 70.0°F, determine the amount of radiant energy delivered to the room by the stove each second if the emissivity is 0.920. (b) If the room is a square with walls that are 8.00 ft high and 25.0 ft wide, determine the R-value needed in the walls and ceiling to maintain the inside temperature at 70.0°F if the outside temperature is 32.0°F. Note that we are ignoring any heat conveyed by the stove via convection and any energy lost through the walls (and windows!) via convection or radiation.

(a)

To determine
Determine the amount of radiant energy delivered to the room by the stove each second.

Explanation

Section1:

To determine : Equivalent of Fahrenheit temperature of room and stove in Kelvin.

Answer: Temperature of room is 294 K and Temperature of stove is 477 K.

Explanation:

Given Info: stove operating temperature is 400°F . Temperature of the room is 70.0°F .

Covert the temperature from Fahrenheit to Celsius.

TC=59(TF32°F)

Substitute Ts for TC and 400°F for TF to calculate stove temperature in Celsius.

Ts=59(400°F32°F)=204°C

Convert Celsius to Kelvin.

TK=273K+TC

Substitute Ts for TK and 204°C for TC to find calculate stove temperature in Kelvin.

Ts=(273+204)K=477K

Thus, stove temperature in Kelvin is 477 K.

Similarly convert room temperature in Kelvin.

Covert the temperature from Fahrenheit scale to Celsius.

TC=59(TF32°F)

Substitute Tr for TC and 70.0°F for TF to calculate room temperature in Celsius.

Tr=59(70.0°F32°F)=21.1°C

Convert Celsius to Kelvin.

TK=273K+TC

Substitute Tr for TK and 21.1°C for TC to find calculate room in Kelvin.

Tr=(273+21.1)K=294K

Thus, temperature of room is 294 K and Temperature of stove is 477 K.

Section2:

To determine : Rate of energy radiated by the stove to the room.

Answer: The amount of radiant energy delivered to the room by the stove each second is 2.03×103W .

Explanation:

Given info: Diameter of cylindrical shape stove is 40.0 cm, length of the stove is 50.0 cm, stove operating temperature is 400°F . Temperature of the room is 70.0°F .

Formula to calculate the net power radiated to the room is,

P=σAse(Ts4tr4)

• P is the net power radiated to the room,
• σ is the Stefan’s –Boltzmann constant,
• Ts is temperature of the stove,
• Tr is temperature of the room,
• e is the emissivity of the stove,
• As is the surface area of the stove,

The total surface area of the cylinder consists of two circular regions on top and bottom and cylindrical side

(b)

To determine
Determine the R value needed in the walls and ceiling to maintain the a temperature.

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