Question

A kettle with 2 litre water at 27${}^{\circ }$ is heated by operating coil heater of power 1 kW. The heat is lost to the atmosphere at constant rate 160 $\frac{\text{J}}{\text{s}}$, when its lid is open. In how much time will water heated to 77${}^{\circ }$ with the lid open? (specific heat of water = 4.2 $kJ/kg{}^{o}C$

• A. 8 min 20 s
• B. 6 min 2 s
• C. 14 min
• D. 7 min

Answer 1

The correct option is A 8 min 20 s

Volume, V = 2 litre = $2\times {10}^{-3}{m}^3$
Energy supplied by heater = $1000J{s}^{-1}$
Heat lost to the atmosphere = $160J{s}^{-1}$
Heat absorbed by water = $1000-160=840J{s}^{-1}$
$\frac{\text{q}}{\text{t}}=\frac{\text{ms}\Delta \text{T}}{\text{t}}=\frac{\text{V}\rho \text{S}\Delta \text{T}}{t}840=\frac{2\times {10}^{-3}\times {10}^3\times 4200\times 50}{\text{t}}\Rightarrow \text{t}=500\text{s}=8\text{min}20\text{s}$