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

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

Tardigrade Admin · Accepted Answer

Heat gained by the water = (Heat supplied by the coil) - (Heat dissipated to environment) ⇒ m c Δ θ=P text Coil t-P text Loss t 4 ⇒ 2 × 4.2 × 103 ×(77-27)=1000 t-160 t ⇒ t=(4.2 × 105/840)=500 s =8 min 20 s

Q. Water of volume 2 litre 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? [Given specific heat of water is 4.2kJ/kg]

8 min 20 s

6 min 2 s

7 min

14 min

Heat gained by the water = (Heat supplied by the coil) - (Heat dissipated to environment) ⇒mcΔθ=PCoil ​t−PLoss ​t 4⇒2×4.2×103×(77−27)=1000t−160t ⇒t=8404.2×105​=500s=8min20s

Heat gained by the water $=$ (Heat supplied by the coil) - (Heat dissipated to environment)
$\Rightarrow m c Δ θ = P_{Coil} t - P_{Loss} t$
$4 \Rightarrow 2 \times 4.2 \times 1 0^{3} \times (77 - 27) = 1000 t - 160 t$
$\Rightarrow t = \frac{4.2 \times 1 0 ^{5}}{840} = 500 s = 8 min 20 s$