Specific heat of water is 40oC . If light of frequency 1.69× 1029 is used to heat 400 g of water from 1.69× 1028 to 2.80× 104 , the number of moles of photons needed will be

Question

Specific heat of water is 40oC . If light of frequency 1.69× 1029 is used to heat 400 g of water from 1.69× 1028 to 2.80× 104 , the number of moles of photons needed will be (A)  2.80× 105  (B)  104Ω  (C)  10Ω  (D)  4.62× 10-2

Tardigrade Admin · Accepted Answer

E=nhv or ms Δ T=nhv n=(ms Δ T/hv) =(400× 4.2× (40-20)/6.625× 10-34× 3× 105) =1.69× 1029 Photons =(1.69× 1029/6.023× 1023) =2.8× 105 Photon/mol

Q. Specific heat of water is $40^{o} C$ . If light of frequency $1.69 \times 10^{29}$ is used to heat 400 g of water from $1.69 \times 10^{28}$ to $2.80 \times 10^{4}$ , the number of moles of photons needed will be

$2.80 \times 10^{5}$

$10^{4} Ω$

$10Ω$

$4.62 \times 10^{- 2}$

$E = nh v$ or $m s Δ T = nh v$ $n = \frac{m s Δ T}{h v}$ $= \frac{400 \times 4.2 \times ( 40 - 20 )}{6.625 \times 10 ^{- 34} \times 3 \times 10 ^{5}}$ $= 1.69 \times 10^{29}$ Photons $= \frac{1.69 \times 10 ^{29}}{6.023 \times 10 ^{23}}$ $= 2.8 \times 10^{5}$ Photon/mol

Q. Specific heat of water is 40oC . If light of frequency 1.69×1029 is used to heat 400 g of water from 1.69×1028 to 2.80×104 , the number of moles of photons needed will be

2.80×105

104Ω

10Ω

4.62×10−2