Physics
Physics
3rd Edition
ISBN: 9781259233616
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 14, Problem 85P

(a)

To determine

The temperature of the leaf.

(a)

Expert Solution
Check Mark

Answer to Problem 85P

The temperature of the leaf is 39°C.

Explanation of Solution

Write the expression for Stefan’s law of radiation.

p=eσAT4

Here, σ is the Stefan constant, e is the emissivity, A is the area, T is the temperature, p is the power absorbed.

Re-arrange the expression to find the temperature.

T=(peσA)14

Substitute ItopA+eσATs4 for p to find the temperature of the leaf.

T=(ItopA+eσATs4eσAb)14

Here, Itop is the intensity at the top surface of the leaf, Ts is the temperature of the surroundings, Ab area of both side leaf.

Conclusion:

Substitute 0.700(9.00×102Wm2) for Itop, 5.00×103m2 for A, 1 for e, 5.670×108Wm2K4 for σ, 25.0°C for Ts and 2(5.00×103m2) for Ab to find the temperature of the leaf.

T=((0.700(9.00×102Wm2))(5.00×103m2)+(1)(5.670×108Wm2K4)(5.00×103m2)((25.0°C+273K)K)4(1)(5.670×108Wm2K4)(2(5.00×103m2)))14=(5.39W(1)(5.670×108Wm2K4)(2(5.00×103m2)))14=312K273K=39°C

(b)

To determine

The power per unit area need to be lost by other methods to keep the temperature at 25.0°C.

(b)

Expert Solution
Check Mark

Answer to Problem 85P

The power per unit area need to be lost by other methods to keep the temperature at 25.0°C is 182W/m2.

Explanation of Solution

The one side of the leaf, the bottom side can be neglected because it absorb and emit in same rate.

Write the expression to find the power per unit area need to be lost.

pab,sunA=praA+pothrA=eσT4+pothrA

Here, pab,sunA is the radiant energy absorbed by the top surface of the leaf, pothrA is the power per unit area need to be lost.

Re-arrange the expression to find the power per unit area need to be lost.

pothrA=pab,sunAeσT4

Conclusion:

Substitute 0.700(9.00×102Wm2) for pab,sunA, 1 for e, 5.670×108Wm2K4 for σ and 25.0°C for Ts to find the t power per unit area need to be lost.

pothrA=0.700(9.00×102Wm2)(1)(5.670×108Wm2K4)((25.0°C+273K)K)4=182W/m2

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Chapter 14 Solutions

Physics

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