Assume that similar quantities of heat are transferred into various masses or mercury and water, resulting in equivalent temperature variations. Mercury's mass is 0.033 times that of water.
Of such eight moons of jupiter, Mercury is the smallest and one that is closest to the Sun. Mercury completes three spins of its axis for each of the two circuits it makes around the Sun, so takes around 96 Earth days. This rotation is exclusive to the solar system and is gravitationally locked.
According to the statement we have Mass of mercury = M(Hg), and
Mass of Water = M(H₂O)
M(cu) / M(H₂O) , heat transfer and mass are related as we have the formula : Q=mcΔT
according to the formula we have , Q(Hg) = M(Hg) C(Hg)ΔT(Hg)...........(1)
for water , Q(H₂O)= M(H₂O) C(H₂O)ΔT(H₂O)..............................(2)
therefore dividing by specific heat in the temperature = C T from both sides .
we Get : [tex]\frac{ Q(Hg)}{CT} =[/tex][tex]\frac{ M(Hg) C(Hg)ΔT(Hg)}{CT}[/tex] , [tex]\frac{Q(H2O) }{CT} =[/tex] [tex]\frac{C(H2O) Δ TH2O}{CT}[/tex]
∴ Q(Hg)=[tex]\frac{Q(Hg)}{Q(Hg)ΔT(Hg)}[/tex] , Q(H₂O)= [tex]\frac{ Q(H2O)}{C(H2O)ΔTH2O)}[/tex]
by solving we get = Q(Hg)/C(Hg)ΔT(Hg)/Q(H₂O) / C(H₂O)ΔT(H₂O)
= Q(Hg)/C(Hg)ΔT(Hg) × C(H₂O)ΔT(H₂O)/Q(H₂O)
= [tex]\frac{C(H2O)}{C(Hg)}[/tex]= M(Hg)/M(H₂O)
specific heat of water = 4184 J/kg
and specific heat of mercury = 139 J/kg
therefore we get 4184/139 = 0.033 ans or 3:100 in ratio
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