Question

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

• A. $150s$
• B. $100s$
• C. $50s$
• D. $200s$

Answer 1

The correct option is A $150s$

Given,

$P=836W$

$\Delta T={40}^0-{10}^0={30}^{0}C$

$1L=1000g$

Heat produced by the heater in time $t$ is,

$Pt=ms\Delta T$

$t=\frac{ms\Delta T}{P}$

$t=\frac{1000\times 4.186\times 30}{836}=150sec$

The correct option is A.