You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Excess Pressure in Bubbles

0.5 kg lemon ...

Question

$0.5$ kg lemon squash at $30^{\circ} C$ is placed in a refrigerator which can remove heat at an average rate of $30 J s^{- 1}$ . How long will it take to cool the lemon sqash to $5^{\circ} C$ ? Specific heat capacity of squash = $4200 J k g - 1 K - 1$

29 min

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

29 s

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

29 min 10 s

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

19 min 10 s

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

Open in App

Solution

The correct option is B 29 min 10 s
Heat released by lemon squash is $0.5 \times 4200 \times (30 - 5) = 52500 J$
So the time required is $52500 / 30 = 1750 s = 29 min 10 sec$

Suggest Corrections

Similar questions

0.5 kg of lemon squash at $30^{\circ} C$ is placed in a refrigerator which can remove heat at an average rate of 30 J $s^{- 1}$ .How long will it take to cool the lemon squash to $5^{\circ} C$ ? Specific heat capacity of squash=4200 J k $g^{- 1} K^{- 1}$ .

0.5 k g

of lemon squash at

30^{o} C

is placed in a refrigerator which can remove heat at an average rate of

30 J s^{- 1}

. How long will it take to cool the lemon squash to

5^{o} C

? Specific heat capacity of squash

= 4200 J k g^{- 1} K^{- 1}

0.5 kg of lemon squash at $30^{0} C$ is placed in a refrigerator which can remove heat at an average rate of 30 $J s^{- 1}$ . How long will it take to cool the lemon squash to $5^{0} C$ ? (Specific heat capacity of squash = 4200 J $K g^{- 1} K^{- 1})$ .

Q. 0.5 kg of lemon squash at 30⁰C is placed in a refrigerator which can remove heat at an average rate of 30 Js^-1. How long will it take to cool the lemon squash to 5⁰C?(Sp. heat capacity of lemon squash = 4200 J kg^{-1 o}C^-1.)

Q. A refrigerator converts

100 g

of water at

20^{\circ} C

to ice at

- 10^{\circ} C

73.5 m i n

. Calculate the average rate of heat extraction in watt. The specific heat capacity of water is

44.2 J g^{- 1} K^{- 1}

, specific latent heat of ice is

336 J g^{- 1}

the specific heat capacity of ice is

2.1 J g^{- 1} K^{- 1}

Join BYJU'S Learning Program

0.5 kg lemon squash at 30∘C is placed in a refrigerator which can remove heat at an average rate of 30Js−1. How long will it take to cool the lemon sqash to 5∘C ? Specific heat capacity of squash = 4200Jkg−1K−1

The correct option is B 29 min 10 sHeat released by lemon squash is 0.5×4200×(30−5)=52500 JSo the time required is 52500/30=1750 s=29min10sec

$0.5$ kg lemon squash at $30^{\circ} C$ is placed in a refrigerator which can remove heat at an average rate of $30 J s^{- 1}$ . How long will it take to cool the lemon sqash to $5^{\circ} C$ ? Specific heat capacity of squash = $4200 J k g - 1 K - 1$

The correct option is B 29 min 10 s
Heat released by lemon squash is $0.5 \times 4200 \times (30 - 5) = 52500 J$
So the time required is $52500 / 30 = 1750 s = 29 min 10 sec$