1. Tardigrade
  2. Question
  3. Physics
  4. Water of volume 2 litre 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? [Given 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 $2$ litre 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 $77^{\circ}C$?
[Given specific heat of water is $4.2 \,kJ/kg$]

BITSATBITSAT 2018

Solution:

Heat gained by the water $=$ (Heat supplied by the coil) - (Heat dissipated to environment)
$\Rightarrow m c \Delta \theta=P_{\text {Coil }} t-P_{\text {Loss }} t$
$4 \Rightarrow 2 \times 4.2 \times 10^{3} \times(77-27)=1000 t-160 t$
$\Rightarrow t=\frac{4.2 \times 10^{5}}{840}=500 s =8 \min 20 s$