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

Byju's Answer

Standard VI

Chemistry

Interconversion of State

Steam at 10...

Question

Steam at $100^{0}$ C is added slowly to $1400$ gm of water at $16^{0} C$ until the temperature of water is raised to $80^{0} .$ The mass of steam required to do this is $(L_{V} = 540 c a l / g m)$

160 gm

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

125 mg

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

250 gm

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

320 gm

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

Open in App

Solution

The correct option is B 160 gm
Let the mass of steam be $x$ .
Heat lost by steam to come to $80^{0} C$ from $100^{0} C$ is
$Q_{l o s t} = x \times 540 + x \times 1 \times (100 - 80) = 560 x$ .....(1)
Heat gained by water $Q_{g a i n e d} = 1400 \times 1 \times (80 - 16) = 1400 \times 64$ ....(2)
Equating (1) and (2)
$560 x = 1400 \times 64$
$∴ x = \frac{140 \times 64}{56} = 160 g m s$

Suggest Corrections

Similar questions

Q. Steam at

100^{o} C

is added slowly to

1400 g m

of water at

16^{o} C

until the temperature of water is raised to

80^{o} C

. The mass of steam required to do this is

(L_{v} = 540 c a l / g m)

Q. The amount of heat required to convert 1 g of ice (specific 0.5 cal at

g^{- 1 o} C^{- 1}

) at

- 10^{0} C

to steam at

100

^{\circ} C

is ___________.

[ Given: Latent heat of ice is

80 C a l / g m,

Latent heat of steam is

540 C a l / g m

, Specific heat of water is

1 C a l / g m / C

]

Q. In an industrial process

10 k g

of water per hour is to be heated from

20^{\circ} C

80^{\circ} C

. To do this steam at

150^{\circ} C

is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at

90^{\circ} C

. How many

k g

of steam is required per hour.
(Specific heat of steam =

1 c a l o r i e

per

g m^{\circ} C

, Latent heat of vaporisation =

540 c a l / g m

4 g m s

of steam at

100^{\circ} C

is added to

20 g m s

of water at

46^{\circ} C

in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation

= 540 c a l / g m

. Specific heat of water

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

Q. 4 gms of steam at

100^{0} C

is added to 20 gms of water at

46^{0} C

in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of steam (in gm) in container at thermal equilibrium is: (Latent heat of vaporisation

=

540 cal/gm. Specific heat of water

= 1 c a l / g m

^{0} C)

Join BYJU'S Learning Program

Steam at 1000C is added slowly to 1400 gm of water at 160C until the temperature of water is raised to 800. The mass of steam required to do this is (LV=540cal/gm)

The correct option is B 160 gmLet the mass of steam be x.Heat lost by steam to come to 800C from 1000C isQlost=x×540+x×1×(100−80)=560x .....(1)Heat gained by water Qgained=1400×1×(80−16)=1400×64 ....(2)Equating (1) and (2)560x=1400×64∴x=140×6456=160gms

Steam at $100^{0}$ C is added slowly to $1400$ gm of water at $16^{0} C$ until the temperature of water is raised to $80^{0} .$ The mass of steam required to do this is $(L_{V} = 540 c a l / g m)$