SATHEE: Properties Of Solids And Liquids Question 109

Properties Of Solids And Liquids Question 109

Question: Water of volume 2 litre in a container is heated with a coil of $1 k W$ 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 $4.2 k J / k g$ ] [IIT-JEE (Screening) 2004]

Options:

A) 8 min 20 s

B) 6 min 2 s

C) 7 min

D) 14 min

Show Answer

Answer:

Correct Answer: A

Solution:

Heat gained by the water = (Heat supplied by the coil) - (Heat dissipated to environment)

therefore $m c Δ θ = P_{C o i l} t - P_{L o s s} t$

therefore $2 \times 4.2 \times 10^{3} \times (77 - 27) = 1000 t - 160 t$

therefore $t = \frac{4.2 \times 10^{5}}{840} = 500 s e c = 8 m i n 20 s e c$

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Properties Of Solids And Liquids Question 109

Question: 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 ] [IIT-JEE (Screening) 2004]

Options:

Answer:

Solution:

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