Estimate: boiling time

Problem

Estimate the time needed to bring $1{,}5\;\text{L}$ of water to the boil on a $1{,}5\;\text{kW}$ hob with efficiency $\sim 70\%$. Then check the estimate with a simple kitchen experiment.

Solution

Setup. The water starts at room temperature ($\sim 20\;^\circ\text{C}$) and must reach $100\;^\circ\text{C}$, so $\Delta T = 80\;\text{K}$. With $1{,}5\;\text{L} \Rightarrow m = 1{,}5\;\text{kg}$ (density $\approx 1\;\text{kg/L}$), the heat needed is: $Q = m\,c\,\Delta T = 1{,}5 \cdot 4186 \cdot 80 = 502320\;\text{J}$

Useful power. Only $70\%$ of the hob’s power reaches the water: $P_\text{useful} = 0{,}70 \cdot 1500 = 1050\;\text{W}$

Time. $t = \frac{Q}{P_\text{useful}} = \frac{502320}{1050} \approx 478\;\text{s}$

$\ev{t \sim 8\;\text{min}}$

Experimental check. Time the actual boiling in the kitchen: a value between $7$ and $10$ minutes confirms the estimate. Discrepancies indicate an actual efficiency different from the assumed $70\%$ (heat lost from the pan, no lid, etc.).

Links

Topics: Thermodynamics Concepts: Specific heat and heat capacity Skills: Fermi estimation