184

You visited us 184 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Physics

Calorimeter

A kettle with...

Question

A kettle with $2$ litre water at $27^{\circ}$ C is heated by operating coil heater of power $1$ kW. The heat is lost to the atmosphere at constant rate $160$ J/sec, when its lid is open. In how much time will water be heated to $77^{\circ}$ C with the lid open'? (specific heat of water = $4.2$ kJ/k g)

8 min 20 sec

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

6 min 2 sec

No worries! We‘ve got your back. Try BYJU‘S free classes today!

14 min

No worries! We‘ve got your back. Try BYJU‘S free classes today!

7 min

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 8 min 20 sec
By the law of conservation of energy. energy given by heater must be equal to the sum of energy gained by water and energy lost from the lid.
$P t = m s$ $Δ θ$ + energy lost
$1000 t = 2 \times 4.2 \times 10^{3} \times 50 + 160 t$
$840 t = 8.4 \times 10^{3} \times 50 = 500 s e c = 8 m i n 20 s e c$

Suggest Corrections

Similar questions

Q. A kettle with

2 L

water at

27^{\circ} C

is heated by operating coil heater power

1 k W

. T h e h e a t i s l o s t t o t h e a t m o s p h e r e a t c o n s t a n t r a t e

160\ J/s

, w h e n i t s l i d i s o p e n . I n h o w m u c h t i m e w i l l w a t e r h e a t e d t o

77^{\circ}C

w i t h t h e l i d o p e n ? (s p e c i f i c h e a t o f w a t e r =

4.2 \ kJ/^{\circ} C - kg$)

Q. Two litres of water at initial temperature of

27^{\circ} C

is heated by a heater of power

1 kW

in a kettle. If the lid of the kettle is open, then heat energy is lost at a constant rate of

160 J/s

. The time in which the temperature will rise from

27^{\circ} C

77^{\circ} C

is (specific heat of water

= 4.2 kJ/kg

Q. Two litre of water at initialled temperature of

27^{o}

C is heated by a heater of power

1

kW. If the lid of kettle is opened, then heat is lost at the constant rate of

160

J/s. Find the time required to raise the temperature of water to

77^{o}

C with the lid open (specific heat of water

4.2

kJ.kg).

Q. Two litre water at initial temperature of

27^{o} C

is heated by a heater of power

1

k W

. If the lid of kettle is opened, then heat is lost at the constant rate of

160

J / s

. Find the time required to raise the temperature of water to

77^{o} C

with the lid open :

(Specific heat of water

4.2

k J

k g

)

Q. Two litre of water at inital temperature of

27^{o} C

is heated by a heater of power

1 k W

. If the lid of kettle is opened, then heat is lost at the constant rate of

160 J / s

. Find, the time required to raise the temperature of water at

77^{o} C

with the lid open (specific of water

4.2 k J / k g

)

Join BYJU'S Learning Program

A kettle with 2 litre water at 27∘C is heated by operating coil heater of power 1 kW. The heat is lost to the atmosphere at constant rate 160 J/sec, when its lid is open. In how much time will water be heated to 77∘C with the lid open'? (specific heat of water = 4.2 kJ/k g)

The correct option is A 8 min 20 secBy the law of conservation of energy. energy given by heater must be equal to the sum of energy gained by water and energy lost from the lid.Pt=ms Δθ + energy lost1000t=2×4.2×103×50+160t840t=8.4×103×50=500sec=8min 20sec