Electric cooker

During which time does an electric cooker with power input P = 500 W and efficiency n = 75% heat water with mass m = 2 kg and temperature t1 = 10°C to the boiling point (t2 = 100°C)?
The specific heat capacity of water is c = 4 180 J/kg/K.

Correct answer:

t =  33.44 min

Step-by-step explanation:

$$\begin{gathered} P=500\text{W} \\ n=75\%=\frac{75}{100}=\frac{3}{4}=0.75 \\ m=2\text{kg} \\ {T}_1=10\text{°C} \\ {T}_2=100\text{°C} \\ c=4180\text{J/kg/K} \\ T=T_2-T_1=100-10=90\text{K} \\ Q=m\cdot c\cdot T=2\cdot 4180\cdot 90=752400\text{J} \\ {P}_2=P\cdot n=500\cdot \frac{3}{4}=\frac{500\cdot 3}{4}=\frac{1500}{4}=375\text{W} \\ Q=P_{2}t \\ {t}_2=\frac{Q}{P_2}=\frac{752400}{375}=\frac{10032}{5}=2006.4\text{s} \\ t=t_2\to \text{min}=t_2:60\text{min}=2006.4:60\text{min}=33.44\text{min}=\frac{836}{25}\text{min}=33.44\text{min} \end{gathered}$$

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