Time taken by a 836; W heater to heat 1; litre of water from 10^{circ} C to 40^{circ} C is

Q: Time taken by a $836\; W$ heater to heat $1\; litre$ of water from $10^{\circ} C$ to $40^{\circ} C$ is

b $W = JH$ $ \Rightarrow $ $P \times t = J \times m\,s\,\Delta \theta $ $ \Rightarrow $ $t = \frac{{J \times m \times s\Delta \theta }}{P}$ (For water $1\, litre$ = $1\,kg$) $ \Rightarrow $ $t = \frac{{4.2 \times 1 \times 1000 \times (40 - 10)}}{{836}} = 150\,sec$ Short Trick : use formula $t = \frac{{4200 \times m \times \Delta \theta }}{P}$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

Time taken by a $836\; W$ heater to heat $1\; litre$ of water from $10^{\circ} C$ to $40^{\circ} C$ is

A$200$
B$150$
C$836$
D$418$

AIEEE 2004, Medium

STD 11 Science
JEE
STD 12 - 3. current electricity
Physics

Download our app
and get started for free

Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*

Download App

Download our appand get started for free

Similar Questions

Download our app
and get started for free