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

Byju's Answer

Standard XII

Chemistry

Heat of Reaction

Steam is pass...

Question

Steam is passed into 54 g of water at $30^{o} C$ till the temperature of the mixture becomes $90^{o} C$ . If the latent heat of steam is $536 c a l g^{- I}$ , the mass of the mixture will be:

80 g

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

60 g

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

50 g

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

24 g

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

Open in App

Solution

The correct option is B 60 g
Heat required to convert water at $30^{o} C$ to $90^{o} C$ $= 54 \times 1 \times (90 - 30) = 54 \times 60 = 3240 c a l$
Let mass of required steam be $m$
So, $m \times 10 + m \times 536 = 3240 \Rightarrow m = \frac{3240}{546} = 5.93$
Required mass of mixture $= 54 + 5.93 = 59.93 g \approx 60 g$

Suggest Corrections

Similar questions

Q. Steam is passed into

56 g

of water at

30^{\circ} C

till the temperature of mixture becomes

90^{\circ} C

. If the latent heat of steam is

536 cal/g

, the mass of the mixture will be:
(Specific heat capacity of water is

1 {cal/g}^{\circ} C

)

2 g

of steam at

100^{\circ} C

is passed into

6 g

of ice at

0^{\circ} C

. If the latent heat of steam and ice are

540 cal/g

and

80 cal/g

respectively and specific heat capacity of water is

1 cal / g^{\circ} C

, then the final temperature of the mixture is

1 g

of steam at

100^{o} C

and equal mass of ice at

0^{o} 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

1 g

of steam at

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. Steam at

100^{\circ} C

is passed into 20 g of water acquires a temperature of

80^{\circ} C

, the mass of water presents will be [Take specific heat of water 1 cal

g^{- 1} c^{- 1}

latent heat of steam =

540 g^{- 1}

]

Join BYJU'S Learning Program

Steam is passed into 54 g of water at 30oC till the temperature of the mixture becomes 90oC. If the latent heat of steam is 536calg−I, the mass of the mixture will be:

The correct option is B 60 gHeat required to convert water at 30oC to 90oC =54×1×(90−30)=54×60=3240calLet mass of required steam be mSo, m×10+m×536=3240⇒m=3240546=5.93Required mass of mixture =54+5.93=59.93g≈60g

Steam is passed into 54 g of water at $30^{o} C$ till the temperature of the mixture becomes $90^{o} C$ . If the latent heat of steam is $536 c a l g^{- I}$ , the mass of the mixture will be: