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Question-227687

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Question Number 227687 by mr W last updated on 14/Feb/26
Commented by fantastic2 last updated on 14/Feb/26
i get +4^0 C
$${i}\:{get}\:+\mathrm{4}^{\mathrm{0}} {C} \\ $$
Commented by fantastic2 last updated on 14/Feb/26
can you pls post some optics related questions(refraction, reflection in spherical mirror)?
$${can}\:{you}\:{pls}\:{post}\:{some}\:{optics}\:{related} \\ $$$${questions}\left({refraction},\:{reflection}\:{in}\:\right. \\ $$$$\left.{spherical}\:{mirror}\right)? \\ $$
Commented by mr W last updated on 14/Feb/26
i have no such questions.
$${i}\:{have}\:{no}\:{such}\:{questions}. \\ $$
Commented by fantastic2 last updated on 15/Feb/26
ok.thanks
$${ok}.{thanks} \\ $$
Answered by fantastic2 last updated on 14/Feb/26
say mass of water in the cup is m and temp is t^0 C and mass of the cold water is λ and temp is θ using t_(final) =((m_1 s_1 t_1 +m_2 s_2 t_2 )/(m_1 s_1 +m_2 s_2 )) we get temperature of cold water after pouring first cup′s water t_1 =((mt+λθ)/(m+λ)) Δt_1 =((m(t−θ))/(m+λ))=10^0 C ...i similarly t_2 =((2mt+λθ)/(λ+2m)) Δt_2 =((mλ(t−θ))/((λ+2m)(m+λ)))=6 ...ii and t_3 =((3mt+λθ)/(λ+3m)) Δt_3 =((mλ(t−θ))/((λ+3m)(λ+2m))) (ii)/(i) ⇒(λ/(2m+λ))=(6/(10)) ⇒λ=3m putting it in (i) ((m(t−θ))/(m+3m))=((t−θ)/4)=10 ∴t−θ=40^0 C so Δt_3 =((3m^2 ×40)/(30m^2 ))=4^0 C
$${say}\:{mass}\:{of}\:{water}\:{in}\:{the}\:{cup}\:{is}\:{m} \\ $$$${and}\:{temp}\:{is}\:{t}^{\mathrm{0}} {C} \\ $$$${and}\:{mass}\:{of}\:{the}\:{cold}\:{water}\:{is}\:\lambda \\ $$$${and}\:{temp}\:{is}\:\theta \\ $$$${using} \\ $$$${t}_{{final}} =\frac{{m}_{\mathrm{1}} {s}_{\mathrm{1}} {t}_{\mathrm{1}} +{m}_{\mathrm{2}} {s}_{\mathrm{2}} {t}_{\mathrm{2}} }{{m}_{\mathrm{1}} {s}_{\mathrm{1}} +{m}_{\mathrm{2}} {s}_{\mathrm{2}} } \\ $$$${we}\:{get} \\ $$$${temperature}\:{of}\:{cold}\:{water}\:{after}\: \\ $$$${pouring}\:{first}\:{cup}'{s}\:{water} \\ $$$${t}_{\mathrm{1}} =\frac{{mt}+\lambda\theta}{{m}+\lambda} \\ $$$$\Delta{t}_{\mathrm{1}} =\frac{{m}\left({t}−\theta\right)}{{m}+\lambda}=\mathrm{10}^{\mathrm{0}} {C}\:…{i} \\ $$$${similarly}\: \\ $$$${t}_{\mathrm{2}} =\frac{\mathrm{2}{mt}+\lambda\theta}{\lambda+\mathrm{2}{m}} \\ $$$$\Delta{t}_{\mathrm{2}} =\frac{{m}\lambda\left({t}−\theta\right)}{\left(\lambda+\mathrm{2}{m}\right)\left({m}+\lambda\right)}=\mathrm{6}\:…{ii} \\ $$$$\:{and} \\ $$$${t}_{\mathrm{3}} =\frac{\mathrm{3}{mt}+\lambda\theta}{\lambda+\mathrm{3}{m}} \\ $$$$\Delta{t}_{\mathrm{3}} =\frac{{m}\lambda\left({t}−\theta\right)}{\left(\lambda+\mathrm{3}{m}\right)\left(\lambda+\mathrm{2}{m}\right)} \\ $$$$\left({ii}\right)/\left({i}\right) \\ $$$$\Rightarrow\frac{\lambda}{\mathrm{2}{m}+\lambda}=\frac{\mathrm{6}}{\mathrm{10}} \\ $$$$\Rightarrow\lambda=\mathrm{3}{m} \\ $$$${putting}\:{it}\:{in}\:\left({i}\right) \\ $$$$\frac{{m}\left({t}−\theta\right)}{{m}+\mathrm{3}{m}}=\frac{{t}−\theta}{\mathrm{4}}=\mathrm{10} \\ $$$$\therefore{t}−\theta=\mathrm{40}^{\mathrm{0}} {C} \\ $$$${so} \\ $$$$\Delta{t}_{\mathrm{3}} =\frac{\mathrm{3}{m}^{\mathrm{2}} ×\mathrm{40}}{\mathrm{30}{m}^{\mathrm{2}} }=\mathrm{4}^{\mathrm{0}} {C} \\ $$
Commented by mr W last updated on 14/Feb/26
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Answered by mr W last updated on 14/Feb/26
say the mass of hot water in each cup is m_2 with temperature T_2 . say the mass of cold water in the container is m_1 with temperature T_1 . T_1 <T_2 . say if we pour hot water with mass m into cold water in container and their temperature becomes T, then m_1 (T−T_1 )=m(T_2 −T) let k=(m_1 /m_2 ), =T_2 −T_1 x=rise of temperature after third pour after first pour: m=m_2 , T=T_1 +10 m_1 (T_1 +10−T_1 )=m_2 (T_2 −T_1 −10) ⇒10k= −10 ...(i) after second pour: m=2m_2 , T=T_1 +10+6 m_1 (T_1 +10+6−T_1 )=2m_2 (T_2 −T_1 −10−6) ⇒ 8k= −16 ...(ii) after third pour: m=3m_2 , T=T_1 +10+6+x m_1 (T_1 +10+6+x−T_1 )=3m_2 (T_2 −T_1 −10−6−x) ⇒ (16+x)k=3( −16−x) ...(iii) from (i) and (ii) we get k=3, =40 put this into (iii): (16+x)3=3(40−16−x) ⇒x=4 ✓ i.e. after third pour the temperature of water in the container will continue to rise by 4°C.
$${say}\:{the}\:{mass}\:{of}\:{hot}\:{water}\:{in}\:{each}\: \\ $$$${cup}\:{is}\:{m}_{\mathrm{2}} \:{with}\:{temperature}\:{T}_{\mathrm{2}} . \\ $$$${say}\:{the}\:{mass}\:{of}\:{cold}\:{water}\:{in}\:{the} \\ $$$${container}\:{is}\:{m}_{\mathrm{1}} \:{with}\:{temperature}\:{T}_{\mathrm{1}} . \\ $$$${T}_{\mathrm{1}} <{T}_{\mathrm{2}} . \\ $$$${say}\:{if}\:{we}\:{pour}\:{hot}\:{water}\:{with}\:{mass} \\ $$$${m}\:{into}\:{cold}\:{water}\:{in}\:{container}\:{and}\: \\ $$$${their}\:{temperature}\:{becomes}\:{T},\:{then} \\ $$$${m}_{\mathrm{1}} \left({T}−{T}_{\mathrm{1}} \right)={m}\left({T}_{\mathrm{2}} −{T}\right) \\ $$$$ \\ $$$${let}\:{k}=\frac{{m}_{\mathrm{1}} }{{m}_{\mathrm{2}} },\:\:={T}_{\mathrm{2}} −{T}_{\mathrm{1}} \\ $$$${x}={rise}\:{of}\:{temperature}\:{after}\:{third}\:{pour} \\ $$$$ \\ $$$${after}\:{first}\:{pour}:\:{m}={m}_{\mathrm{2}} ,\:{T}={T}_{\mathrm{1}} +\mathrm{10} \\ $$$${m}_{\mathrm{1}} \left({T}_{\mathrm{1}} +\mathrm{10}−{T}_{\mathrm{1}} \right)={m}_{\mathrm{2}} \left({T}_{\mathrm{2}} −{T}_{\mathrm{1}} −\mathrm{10}\right) \\ $$$$\Rightarrow\mathrm{10}{k}=\:−\mathrm{10}\:\:\:…\left({i}\right) \\ $$$${after}\:{second}\:{pour}:\:{m}=\mathrm{2}{m}_{\mathrm{2}} ,\:{T}={T}_{\mathrm{1}} +\mathrm{10}+\mathrm{6} \\ $$$${m}_{\mathrm{1}} \left({T}_{\mathrm{1}} +\mathrm{10}+\mathrm{6}−{T}_{\mathrm{1}} \right)=\mathrm{2}{m}_{\mathrm{2}} \left({T}_{\mathrm{2}} −{T}_{\mathrm{1}} −\mathrm{10}−\mathrm{6}\right) \\ $$$$\Rightarrow\:\mathrm{8}{k}=\:−\mathrm{16}\:\:\:…\left({ii}\right) \\ $$$${after}\:{third}\:{pour}:\:{m}=\mathrm{3}{m}_{\mathrm{2}} ,\:{T}={T}_{\mathrm{1}} +\mathrm{10}+\mathrm{6}+{x} \\ $$$${m}_{\mathrm{1}} \left({T}_{\mathrm{1}} +\mathrm{10}+\mathrm{6}+{x}−{T}_{\mathrm{1}} \right)=\mathrm{3}{m}_{\mathrm{2}} \left({T}_{\mathrm{2}} −{T}_{\mathrm{1}} −\mathrm{10}−\mathrm{6}−{x}\right) \\ $$$$\Rightarrow\:\left(\mathrm{16}+{x}\right){k}=\mathrm{3}\left(\:−\mathrm{16}−{x}\right)\:\:\:…\left({iii}\right) \\ $$$$ \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{we}\:{get} \\ $$$${k}=\mathrm{3},\:\:=\mathrm{40} \\ $$$${put}\:{this}\:{into}\:\left({iii}\right): \\ $$$$\left(\mathrm{16}+{x}\right)\mathrm{3}=\mathrm{3}\left(\mathrm{40}−\mathrm{16}−{x}\right) \\ $$$$\Rightarrow{x}=\mathrm{4}\:\checkmark \\ $$$${i}.{e}.\:{after}\:{third}\:{pour}\:{the}\:{temperature} \\ $$$${of}\:{water}\:{in}\:{the}\:{container}\:{will} \\ $$$${continue}\:{to}\:{rise}\:{by}\:\mathrm{4}°{C}. \\ $$

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