Question

An electric kettle was marked 500 W, 230 V, and was found to take 15 minutes to bring 1 kg of water at 15°C to boiling point. Determine the heat efficiency of the kettle. 
1. 70%
2. 75%
3. 85%
4. None of these

Answer 1

Correct Answer - Option 4 : None of these

Concept:

Heat efficiency of kettle = (Total heat output) / (Total heat input)

H = MSΔT

M is the mass

S is the specific heat

ΔT is the temperature rise

Calculation:

Given marked power setting, P = 500 watt

t = 15 min = 900 seconds

ΔT = 85°C

S = 4200 J/kg °C

M = 1 kg

Input heat = Hinput = p x t = (500 × 900) Joules

Output heat = Houtput = MSΔT = (1 × 4200 × 85) Joules

Efficiency (η) is given as,

\(\large{η=\dfrac{H_{output}\times100}{H_{input}}}=\dfrac{4200\times85\times 100}{500\times900}=79.3\%\)=Houtput×100Hinput=4200×85×100500×900=79.3%