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

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 ] (A) 8 min 20 s (B) 6 min 2 s (C) 7 min (D) 14 min

Tardigrade Admin · Accepted Answer

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° C to 77° C is m s Δ θ. Hence, the required time t =(m s Δ θ/ text rate by which energy is gained by water ) =((2)(4.2 × 103)(50)/840)=500 s =8 min 20 s

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

8 min 20 s

6 min 2 s

7 min

14 min

Energy gained by water (in 1s ) = energy supplied - energy lost =1000J−160J=840J Total heat required to raise the temperature of water from 27∘C to 77∘C is msΔθ. Hence, the required time t= rate by which energy is gained by water msΔθ​ =840(2)(4.2×103)(50)​=500s=8min20s

Q. Water of volume $2 L$ in a container is heated with a coil of $1 kW$ at $2 7^{\circ} C$ . The lid of the container is open and energy dissipates at rate of $160 J / s$ . In how much time temperature will rise from $2 7^{\circ} C$ to $7 7^{\circ} C$ ?
[Specific heat of water is $4.2 k J / k g$ ]