Question

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Constants: R=0.08206 L atm K'mol = 8.31451 J K 00 L atm K 'mol = 8.31451 J K molt mol' c = 2998 x 108 msh-6.626 x 10 c = 2.998 x 10 JS A sample contains 17.80 x 1021 molecules of liquid water at an initial temperature of The sample absorbs 1.211 x 109 photons of 2081 nm IR (ie heat). ater at an initial temperature of the water is 25.00°C. Note: the specific heat of liquid water = 4.180 J/g °C a) How much energy (in J) is found with a single 2081 nm photon? b) How much energy (in J) would be found with 1.211 x 10"photons of the same wavelength as part a? c) What is the mass of water? d) What is the final temperature of the water after it absorbs all the photons?

Answer 1

Given - 1. sol". Constant: R=0.08206 Latu Kimol" - 0.31451 JK molt c= 2.9988108 mst, h = 6.626 X 10 34 Jos A stample contains 17.80 4102 molecules of liquid water at an initial temperature of the water is 25.000. The sample absombs 1.211 X70/9 photons of 2001 1m IR lie heaty. specific heat of liquid water = 4180 T/gee a) To find- Energy found with a single 2081 nm photon formula Used R E ho E = Energy h: Plank's castaust I nom = 1x100m zodium= 2081810 m c= speed of light d = wavelength putting all values in equations (i) 6.626 X10-34 x 2.9981100 Joule Here, E 2001 X109 E = 9.54578 x 10-20 g Therefore, Energy found with a single 2001 nm photon is 9154578 X1020]

(6) To find How much euergy (in J) would be found with 1.211 X 109 puotans of the same wavelength as part a? Energy of a photon = 9.54578 110 20 7 Energy of 1.211 X1019 puotams = 9.54578x1020 x 1.2118/0 = 1.156 ] (C) To find - mass of water No.molecules : No.of moles X NA -C) NA = 6.02281623 (Avogadro number) Given molecules ay liquid water: 1700x10e putting this value in equation is 17.80X10216.022 x 1023 x No. of moles No. of moles: 2.9568102 mol Cliven mass (gram) ang No. of moles: Molar mass molar mass of M20W) = 18.01528 g fuel By using equation (iii) we get 2.956x10² MHzole (mass of the liquid) 18.01520 mass of Mo liquid = 5.3253x107 gram

Therefore, mass of water is 5.32538/olgsam. (d) To find Final temperature of the water after it absorbs all the puotans. fonmula ureda q=meat q- Energy absorb me mass of H90 (here) c- specific heat of liquid water OT = (1-77) °C 7 - Final Temperature Tin Juitial Temperature. Using all the value we find in this question. q= 9.54578716-20 x 1.21181019 9 = 11565 m = 5.3253x10 gram c= 4.180 J/gcc Ti= 25°C 1.156 = 5.3253 X10'x 4.180x (17-25°C) T: 25 °C + 0.519°C T = 25.519°C Therefore, final temperature of water is 25:519°