Worked example — Copper pot in water

Problem

A $500$ g copper pot at $200\ ^{\circ}$C is plunged into $1$ kg of water at $20\ ^{\circ}$C. What is the equilibrium temperature? Data: $c_{\text{rame}}=385\ \mathrm{J/(kg\cdot K)}$, $c_{\text{acqua}}=4186\ \mathrm{J/(kg\cdot K)}$.

Solution

Setup. Isolated system: heat released by the copper plus heat absorbed by the water equals zero. $m_{Cu}\,c_{Cu}\,(T_e-200)+m_{H_2O}\,c_{H_2O}\,(T_e-20)=0$ Substitution (masses in kg): $0{,}5\cdot 385\cdot(T_e-200)+1\cdot 4186\cdot(T_e-20)=0$ $192{,}5\,(T_e-200)+4186\,(T_e-20)=0$ $4378{,}5\,T_e=192{,}5\cdot200+4186\cdot20=38500+83720=122220$ $\ev{T_e\approx 27{,}9\ ^{\circ}\mathrm{C}}$ The water, with its enormous specific heat, “wins” over the copper: the temperature rises by only a few degrees.

Links

Topics: Termologia Concepts: Calore specifico e capacità termica Skills: Equazione calorimetrica Objects: Calorimetro