Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Current Electricity
0 votes

Time taken by a $836\: W$ heater to heat one liter of water from $10^{\circ}C\: to\: 40^{\circ}C$ is

$\begin {array} {1 1} (A)\;50s & \quad (B)\;100s \\ (C)\;150s & \quad (D)\;200s \end {array}$


Can you answer this question?

1 Answer

0 votes
Let $t$ be the time taken
Then heat energy in calories is given by
$ \large\frac{836 \times t}{4.2}$ $= 1000 \times 1 \times (40 – 10)$ [using $Q = mst$ (from heat transfer)]
$\Rightarrow t = 150s$
Ans : (C)


answered Feb 16, 2014 by thanvigandhi_1
edited Sep 10, 2014 by thagee.vedartham

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App