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

Byju's Answer

Standard X

Physics

Latent Heat

A steam engin...

Question

A steam engine intakes $50 g$ of steam at $100^{o} C$ per minute and cools it down to $20^{o} C$ . If latent heat of vaporization of steam is $540 c a l g^{- 1}$ , then the heat rejected by the steam engine per minute is $10^{3} c a l$ .
(Given : specific heat capacity of water : $1 c a l / g^{o} C$ )

Open in App

Solution

Heat rejected by steam = Latent heat + specifc heat to bring down temperature to $20^{o} C$

$Δ Q_{r e j} = m L + m s Δ T$
$Δ Q_{r e j} = 50 \times 540 + 50 \times 1 \times (100 - 20)$
$= 50 \times [540 + 80]$
$= 50 \times 620$
$= 31000 c a l$
$= 31 \times 10^{3} c a l$

Suggest Corrections

Similar questions

Q. Calculate the mass of steam at

100^{\circ} C

needed to just melt 320 g of ice at

- 10^{\circ} C

. The specific heat of ice

= 0.5 c a l g^{- 1}^{\circ} C^{- 1}

, latent heat of vaporization of steam

= 540 c a l g^{- 1},

specific heat of fusion of ice

= 80 c a l g^{- 1}

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

Q. Ice at

0^{\circ} C

is added to

1 g

of steam at

100^{\circ} C

. For

x g

of ice added, the temperature of steam reduces to

0^{\circ} C

. Find the value of

x

.
(Latent heat of ice

= 80 cal/g

and latent heat of steam

= 540 cal/g

. Also, specific heat of water

= 1 cal/ kg K

)

Q. 400 g of ice at 253 K is mixed with 0.05 kg of steam at

100^{o} C

. Latent heat of vaporisation of steam

= 540 c a l g^{- 1}

. Latent heat of fusion of ice

= 80 c a l g^{- 1}

. Specific heat of ice

= 0.5 c a l g^{- 1}^{o} C^{- 1}

. Find the resultant temperature of the mixture.

Q. Work done in converting one gram of ice at

- 10^{\circ} C

into steam at

100^{\circ} C

is -
[Given, specific heat of ice

c_{i} = 0.5 {cal g}^{- 1}^{\circ} C^{- 1}

, specific heat of water

c_{w} = 1 {cal g}^{- 1}^{\circ} C^{- 1}

, latent heat of fusion of ice

L_{f} = 80 cal/g

, & latent heat of vaporization of steam

L_{v} = 540 cal/g

]

Join BYJU'S Learning Program

A steam engine intakes 50 g of steam at 100oC per minute and cools it down to 20oC. If latent heat of vaporization of steam is 540 cal g−1, then the heat rejected by the steam engine per minute is 103 cal. (Given : specific heat capacity of water : 1 cal/g oC)

Heat rejected by steam = Latent heat + specifc heat to bring down temperature to 20oC ΔQrej=mL+msΔT ΔQrej=50×540+50×1×(100−20) =50×[540+80] =50×620 =31000 cal =31×103 cal

Heat rejected by steam = Latent heat + specifc heat to bring down temperature to $20^{o} C$

$Δ Q_{r e j} = m L + m s Δ T$
$Δ Q_{r e j} = 50 \times 540 + 50 \times 1 \times (100 - 20)$
$= 50 \times [540 + 80]$
$= 50 \times 620$
$= 31000 c a l$
$= 31 \times 10^{3} c a l$