Water of volume 2 litre in a container is heated with a coil of 1 kW at . The lid of the container is open and energy dissipates at rate of 160 J/s. In how much time temperature will rise from to [Given specific heat of water is ] [2005]

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 $4.2 kJ / kg$ ] [2005]

Smart Achievers · Accepted Answer

(3)

As shown in the figure, the net heat gained by the water to raise its temperature

$= (1000 - 160) = 840 J/s$

Now, the heat required to raise the temperature of water from $27 ° C$ to $77 ° C$

$Q = m c Δ t = 2 \times 4200 \times 50 J$

Hence the time required to gain $Q$ amount of heat

$t = \frac{Q}{840} = \frac{2 \times 4200 \times 50}{840} = 500 s = 8 min 20 s$

Solution

Q.
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 $4.2 kJ / kg$ ] [2005]

1 7 min

2 6 min 2s

3 8 min 20s

4 14 min

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