CameraIcon

SearchIcon

MyQuestionIcon

2

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Principle of Calorimetry

The final tem...

Question

The final temperature, when 10 g of steam at $100^{o} C$ is passed into an ice block of mass 100 g
$(L_{s t e a m} = 540 c a l / g, L_{i c e} = 80 c a l / g; S_{w a t e r} = 1 c a l / g^{o} C)$ is

A

0^{0} C

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

B

{15.7}^{o} C

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

C

{16.9}^{o} C

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

D

{20.4}^{o} C

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

Open in App

Solution

The correct option is A $0^{0} C$
As the man of steam is for small compared to ice. Lets assume that final temp be $0^{o} C$ .
If
$Δ H$ released $< Δ H$ melting
Assumption is right else wrong
So,
$Δ H$ melting $= m_{i c e} \times L_{i c e}$
$Δ H$ released $= m_{s t e a m} \times L_{s t e a m}$
$+ (m_{s t e a m}) (S_{w a t e r}) | Δ T |$
$= (10) (540) + 10 \times 1 \times 100$
$= 5500 c a l$
As $Δ H$ released $Δ H$ melting, assumption is right.

Suggest Corrections

thumbs-up

0

Similar questions

Q. In an insulated vessel,

0.05 k g

373 K

0.45 k g

253 K

are mixed. Then, find the final temperature of the mixture.
Given,

L_{f u s i o n} = 80 c a l / g = 336 J / g

L_{v a p o r i s a t i o n} = 540 c a l / g = 2268 J / g

S_{i c e} = 2100 J / k g K = 0.5 c a l / g K

S_{w a t e r} = 4200 J / k g K = 1 c a l / g K

5 g

100^{\circ} C

6 g

0^{\circ} C

. If the latent heat of steam and ice are

540 cal/g

80 cal/g

respectively and specific heat capacity of water is

1 cal / g^{\circ} C

, then the final temperature of the mixture is

Q. In an insulated vessel, 0.05 kg steam at 373 K and 0.45 kg of ice at 253 K are mixed. Then, find the final temperature of the mixture.
Given,

L_{f u s i o n} = 80 c a l / g = 336 J / g,

L_{v a p o r i z a t i o n} = 540 c a l / g = 2268 J / g,

S_{i c e} = 2100 J / k g K = 0.5 c a l / g K

S_{w a t e r} = 4200 J / k g K = 1 c a l / g K

1 g

100^{\circ} C

and an equal mass of ice at

0^{\circ} C

are mixed. The temperature of the mixture in steady state will be : (latent heat of steam

= 540 c a l / g

, latent heat of ice

= 80 c a l / g

, specific heat of water

= 1 c a l / g^{\circ} C

)

Q. The heat required (in

c a l

) to change 10 g ice at

0^{o} C

100^{o} C

is:
[ Heat of fusion and heat of vaporization for

H_{2} O

80 c a l / g

540 c a l / g

respectively. Specific heat of water is

1 c a l / g

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Principle of calorimetry

PHYSICS

Watch in App

explore_more_icon

Explore more

Principle of Calorimetry

Standard X Physics

Join BYJU'S Learning Program

CrossIcon