A beaker contains 200 gm of water- The heat capacity of the beaker is equal to that of 20 gm of water- The initial temperature of water in the beaker is 20- C- If 440 gm of hot water at 92- C is poured in it- the final temperature neglecting radiation loss will be nearest to

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

Byju's Answer

Standard XII

Physics

Calorimetry

A beaker cont...

Question

A beaker contains 200 gm of water. The heat capacity of the beaker is equal to that of 20 gm of water. The
initial temperature of water in the beaker is $20^{\circ} C .$ If 440 gm of hot water at $92^{\circ} C$ is poured in it, the final
temperature (neglecting radiation loss) will be nearest to

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

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

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

Open in App

Solution

The correct option is B
Heat lost by hot water = Heat gained by cold water in beaker + Heat absorbed by beaker
$\Rightarrow 440 (92 - T) = 200 \times (T - 20) + 20 \times (T - 20) \Rightarrow T = 68^{\circ} C$

Suggest Corrections

Similar questions

Q. A beaker contains 200 gm of water. The heat capacity of the beaker is equal to that of 20 gm of water. The
initial temperature of water in the beaker is

20^{\circ} C .

If 440 gm of hot water at

92^{\circ} C

is poured in it, the final
temperature (neglecting radiation loss) will be nearest to

Q. A beaker contains

200

g of water.The heat capacity of beaker is equal to that

20

g of water.The initial temperature of water in the beaker is

20^{\circ} C

. If

440

g of hot water at

92^{\circ} C

is poured in the final temperature,neglecting radiation loss, will be

Q. A beaker contains 200 g of water. The heat capacity of the beaker is equal to that of 20 g water. The initial temperature of water in the beaker is

20^{o} C

. If 440 g of hot water at

92^{o} C

is poured in, the final temperature (neglecting radiation loss) will be nearly :

A beaker contains $200 g m$ of water. The heat capacity of the beaker is equal to that of $20 g m$ of water. The initial temperature of water in the beaker is $20^{\circ} C .$ If $440 g m$ of hot water at $92^{\circ} C$ is poured into it, the final temperature od the system is

Q. A vessel contains

110 g

of water. The heat capacity of the vessel is equal to that of

10 g

of water. The initial temperature of water in the vessel is

10^{\circ} C

. While the water and vessel are in thermal equilibrium,

220 g

of hot water at

70^{\circ} C

is poured into the vessel. Find the final temperature of the vessel + water system (neglecting radiation loss).

Join BYJU'S Learning Program

A beaker contains 200 gm of water. The heat capacity of the beaker is equal to that of 20 gm of water. The initial temperature of water in the beaker is 20∘C. If 440 gm of hot water at 92∘C is poured in it, the final temperature (neglecting radiation loss) will be nearest to

The correct option is B Heat lost by hot water = Heat gained by cold water in beaker + Heat absorbed by beaker ⇒440(92−T)=200×(T−20)+20×(T−20)⇒T=68∘C

A beaker contains 200 gm of water. The heat capacity of the beaker is equal to that of 20 gm of water. The
initial temperature of water in the beaker is $20^{\circ} C .$ If 440 gm of hot water at $92^{\circ} C$ is poured in it, the final
temperature (neglecting radiation loss) will be nearest to

The correct option is B
Heat lost by hot water = Heat gained by cold water in beaker + Heat absorbed by beaker
$\Rightarrow 440 (92 - T) = 200 \times (T - 20) + 20 \times (T - 20) \Rightarrow T = 68^{\circ} C$