Heating water

Problem

How much energy is needed to heat $0{,}50\;\text{kg}$ of water from $20\;^\circ\text{C}$ to $60\;^\circ\text{C}$? Use $c_\text{acqua} = 4186\;\text{J/(kg·K)}$.

Solution

Setup. No change of state: the energy is only needed to raise the temperature. The calorimetric equation applies: $Q = m\,c\,\Delta T$ with $m = 0{,}50\;\text{kg}$, $c = 4186\;\text{J/(kg·K)}$ and $\Delta T = 60 - 20 = 40\;\text{K}$.

Numerical calculation. $Q = 0{,}50 \cdot 4186 \cdot 40 = 83720\;\text{J}$

$\ev{Q \approx 8{,}4\cdot 10^4\;\text{J}}$

Links

Topics: Thermodynamics Concepts: Specific heat and heat capacity Skills: Calorimetric equation