Iced Water

Classical Mechanics Level 4

At a stifling hot night, I had decided to make myself a jar of cold water for my own enjoyment. The jar of water was initially at room temperature and cooled by adding ice cubes into it. However, because I did not want to waste any of my precious ice cubes, I added the subsequent ice cube if and only if the previous one had melted. Assuming the jar was a cylinder and there was no heat loss/gain to/from the surroundings, it is given that: c i c e = 2108 J kg X − 1 K X − 1 c w a t e r = 4187 J kg X − 1 K X − 1 l f u s i o n = 334 k J kg X − 1 ρ i c e = 920 k g m X − 3 ρ w a t e r = 1000 k g m X − 3 l c u b e = 1 c m r j a r = 3 c m h w a t e r = 8 c m θ i c e = 270 K θ w a t e r = 300 K θ f u s i o n = 273 K \begin{aligned} { c }_{ ice } & = 2108\space \ce{ J{ kg }^{ -1 }{ K }^{ -1 } } \\ { c }_{ water } & = 4187\space \ce{ J{ kg }^{ -1 }{ K }^{ -1 } } \\ { { l }_{ fusion } } & = 334\space \ce{ kJ{ kg }^{ -1 } } \\ { \rho }_{ ice } & = 920\space \ce{ kg{ m }^{ -3 } } \\ { \rho }_{ water } & = 1000\space \ce{ kg{ m }^{ -3 } } \\ { l }_{ cube } & = 1\space \ce{ cm } \\ { r }_{ jar } & = 3\space \ce{ cm } \\ { h }_{ water } & = 8\space \ce{ cm } \\ { \theta }_{ ice } & = 270\space \ce{ K } \\ { \theta }_{ water } & = 300\space \ce{ K } \\ { \theta }_{ fusion } & = 273\space \ce{ K } \end{aligned} How many ice cubes would I add?


The answer is 82.

Gandoff Tan by Gandoff Tan , 19, Singapore

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1 solution

Gandoff Tan
May 2, 2019

Let "i" = "ice", "w" = "water", "f" = "fusion", "j" = "jar".

Q i = Q w ( m i c i Δ θ i + m i l f ) n i = m w c w Δ θ w n i = ⌈ ρ w V w c w ( θ w − θ f ) ρ i V i ( c i ( θ f − θ i ) + l f ) ⌉ = ⌈ ρ w π r j 2 h w c w ( θ w − θ f ) ρ i l i 3 ( c i ( θ f − θ i ) + l f ) ⌉ = ⌈ ( 1000 ) ( 0.03 ) 2 ( 0.08 ) ( 4187 ) ( 300 − 273 ) ( 920 ) ( 0.01 ) 3 ( ( 2108 ) ( 273 − 270 ) + 334000 ) π ⌉ = ⌈ 81.6711535519 … ⌉ n i = 82 \begin{aligned} \mathsf{Q_ i}& \mathsf{=Q_w}\\ \mathsf{(m_ic_i\Delta\theta_i+m_il_f)n_i}& \mathsf{=m_wc_w\Delta\theta_w}\\ \mathsf{n_i}& \mathsf{=\left\lceil\frac{\rho_wV_wc_w(\theta_w-\theta_f)}{\rho_iV_i(c_i(\theta_f-\theta_i)+l_f)} \right\rceil}\\& \mathsf{=\left\lceil\frac{\rho_w\pi r_j^2h_wc_w(\theta_w-\theta_f)}{ { \rho }_{ i }{ { l }_{ i } }^{ 3 }({ c }_{ i }({ \theta }_{ f }-{ \theta }_{ i })+{ l }_{ f }) } \right\rceil } \\ \quad & \mathsf{= \left\lceil \frac { (1000){ (0.03) }^{ 2 }(0.08)(4187)(300-273) }{ (920){ (0.01) }^{ 3 }((2108)(273-270)+334000) } \pi \right\rceil} \\ \quad & \mathsf{= \left\lceil 81.6711535519\dots \right\rceil} \\\mathsf{ { n }_{ i }} &\mathsf{ =} \boxed{ \mathsf{82} } \end{aligned}

1 Helpful 2 Interesting 0 Brilliant 4 Confused

At what temp would you find it cold enough?

Hudson Snow - 2 years, 1 month ago

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θ f u s i o n \theta_{fusion}

Gandoff Tan - 2 years, 1 month ago

apparently 273K

Aaron Hu - 1 year, 11 months ago

It would just be better if you put more at the same time....

Filip Jerleković - 1 year, 10 months ago

idont even know whats going on here

Tor Park - 1 year, 9 months ago

Hey Gordon, I'd say remove the 30 deg C in the original problem or change that to 27 deg C. Or is that meant to be the difference in temperature between the water and the ice?

Siddharth Gopujkar - 1 year, 8 months ago

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