Worked example — Ice cube in tea

Problem

We drop a $20$ g ice cube, initially at $-5\ ^{\circ}$C, into $200$ g of tea (take it as pure water) at $80\ ^{\circ}$C. What is the final temperature, assuming all the ice melts? Data: $c_{\text{ghiaccio}}=2090$, $c_{\text{acqua}}=4186\ \mathrm{J/(kg\cdot K)}$; $L_f=3{,}34\cdot10^{5}\ \mathrm{J/kg}$.

Solution

Setup. The tea releases heat; the ice absorbs it in three steps: heating up to $0\ ^{\circ}$C, melting, heating the meltwater up to $T_e$. Heat released by the tea: $Q_{ced}=0{,}2\cdot 4186\cdot(80-T_e)$. Heat absorbed by the ice: $Q_1=0{,}02\cdot 2090\cdot 5=209\ \mathrm{J},\quad Q_2=0{,}02\cdot 3{,}34\cdot10^{5}=6680\ \mathrm{J},\quad Q_3=0{,}02\cdot 4186\cdot T_e$ Balance $Q_{ced}=Q_1+Q_2+Q_3$: $837{,}2\,(80-T_e)=209+6680+83{,}72\,T_e$ $66976-837{,}2\,T_e=6889+83{,}72\,T_e$ $60087=920{,}9\,T_e$ $\ev{T_e\approx 65{,}2\ ^{\circ}\mathrm{C}}$ The tea cools by about 15 degrees, mostly because of the latent heat of melting of the ice.

Links

Topics: Termologia Concepts: Calore latente e cambiamenti di stato Skills: Equazione calorimetrica