Physics - Elasticity Question with Solution | TestHub

To find the time taken, we equate the heat supplied by the heater to the heat absorbed by the water.

Given:

Power of heater $P=836$ W

Volume of water $V=1$ liter $=1000$ cm³

Mass of water $m=1000$ g $=1$ kg (since density of water is $1$ g/cm³)

Initial temperature $T_1=10^{\circ }$C

Final temperature $T_2=40^{\circ }$C

Specific heat capacity of water $c=4.2$ J/g${}^{\circ }$C $=4200$ J/kg${}^{\circ }$C

Heat absorbed by water $Q=mc\Delta T$

$Q=(1000g)(4.2J/g^{\circ }C)(40^{\circ }C-10^{\circ }C)$

$Q=(1000)(4.2)(30)$ J

$Q=126000$ J

Heat supplied by heater $Q=P\times t$

$126000J=(836W)\times t$

$t=\frac{126000}{836}\approx 150.7s$

Rounding to the nearest option, $t=150$ s.

The final answer is $\boxed{150s}$.