You visited us 6 times! Enjoying our articles? Unlock Full Access!

Interconversion of State

Steam at 10...

Question

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

160 g m

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

125 g m

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

250 g m

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

320 g m

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

Open in App

Solution

The correct option is A $160 g m$
Heat supplied to water is $Q_{1} = 1400 \times (80 - 16)$
Now this heat supplied to water will be released by condensation of vapor to liquid.
also, $Q_{1} = m (L_{v a p} + 1 \times (100 - 80))$
so, $m = 160 g m$

Suggest Corrections

Similar questions

100^{0}

1400

16^{0} C

80^{0} .

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

100 g

24^{0} C

90^{0} C

100^{o} C

20 g

10^{o} C

80^{o} C

10 k g

20^{\circ} C

80^{\circ} C

150^{\circ} C

90^{\circ} C

k g

1 c a l o r i e

g m^{\circ} C

540 c a l / g m

- 5^{o} C

100^{o} C

1 c a l g^{- 1}^{o} C^{- 1}

0^{o} C

100^{o} C

Join BYJU'S Learning Program