1. Tardigrade
  2. Question
  3. Physics
  4. Water of volume 2 L in a container is heated with a coil of 1 kW at 27° C. The lid of the container is open and energy dissipates at rate of 160 J / s. In how much time temperature will rise from 27° C to 77° C ? [Specific heat of water is 4.2 kJ / kg ]

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Water of volume $2\, L$ in a container is heated with a coil of $1 \,kW$ at $27^{\circ} C$. The lid of the container is open and energy dissipates at rate of $160\, J / s$. In how much time temperature will rise from $27^{\circ} C$ to $77^{\circ} C$ ?
[Specific heat of water is $4.2\, kJ / kg$ ]

IIT JEEIIT JEE 2005Thermodynamics

Solution:

Energy gained by water (in $1 \,s$ )
$=$ energy supplied - energy lost $=1000 \,J -160\, J =840\, J$
Total heat required to raise the temperature of water from
$27^{\circ} C$ to $77^{\circ} C$ is $m s \Delta \theta$.
Hence, the required time
$t =\frac{m s \Delta \theta}{\text { rate by which energy is gained by water }} $
$=\frac{(2)\left(4.2 \times 10^{3}\right)(50)}{840}=500\, s =8 \min 20 \,s$