Class Management Help Ch. 22 Homework, pt. 2 Begin Date: 3/12/2018 2:00:00 PM Du

Question

Class Management Help Ch. 22 Homework, pt. 2 Begin Date: 3/12/2018 2:00:00 PM Due Date: 3/14/2018 5:00:00 PM End Date: 3/1 (25%) Problem 4: Assume the radiation from a heat lamp is monochromatic, with a w avelength of 1.5 m 25% Part a) If the lamp's 180.7-W output is focused on a person's shoulder, over a circular area 24 cm in diameter, what is the intensity in kilowatts per square meter? Grade Summary Deductions Potential 100% 1- 0% sinO cotanOasin0 tan() |T1(1)171819 acosO cosO atan0 acotan inh0 o Degrees Radians HOM Submissions Attempts remaining: 6 (4% per attempt) detailed view 4 5 6 coshO tanh cotanh 0 END BACKSPACE DELCLEAR Submit Hint Feedback I give up Hints: 1 deduction per hint. Hints remaining:1 Feedback: 1% deduction per feedback. 25% Part (b) what is the peak electric field strength, in kilovolts per meter? 25% Part (c) Find the peak rnagnetic field strength, in microtesla. 25% Part (d) How long, in seconds, will it take to increase the temperature of the 42-kg shoulder by 200°C, assuming that the shoulder absorbs all the radiation from the lamp and given that its specific heat is 3.47 x 103 J/(kg·°C)? All content © 2018 Expert TA, LLC

Explanation / Answer

Given,

lambda = 1.5 um

a)P = 180.7 W ; D = 24 cm

We know that,

I = P/A = 180.7/[3.14 x 0.12^2] = 3.996 x 10^3 W/m^2

Hence, I = 3.996 x 10^3 W/m^2

b)We know that, intensity and electric field are related as:

E = sqrt(2I/ce0)

E = sqrt[(2 x 3.996 x 10^3)/(3 x 10^8 x 8.85 x 10^-12)] = 1.734 x 10^3 N/C

Hence, E = 1.734 x 10^3 V/m

c)We know that

c = E/B => B = E/c

B = 1.734 x 10^3/(3 x 10^8) = 5.78 x 10^-6 T

B = 5.78 x 10^-6 T

d)We know that,

Q = m C (T2 - T1)

Also, P = Q/t => Q = Pt

pt = m c (T2 - T1)

t = m C (T2 - T1)/P

t = 4.2 x 3.47 x 10^3 x 2/180.7 = 161.31 sec

Hence, t = 161.31 s

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