Work done in converting one gram of ice at -10-C into steam at 100-C is -Given- specific heat of ice ci -0-5 cal g-1 -C-1- specific heat of water cw -1 cal g-1 -C-1- latent heat of fusion of ice Lf-80 cal-g- latent heat of vaporization of steam Lv-540 cal-g-

Pictorial representation of conversion of ice at 10^∘C into steam at 100^∘C is given below (c_i= specific heat of ice, c_w=specific heat of water) Let Q be t ...

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

Byju's Answer

Standard XI

Physics

Specific Heat Capacity

Work done in ...

Question

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 $L_{v} = 540 cal/g$ ]

725 J

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

3045 J

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

5023 J

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

1128 J

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

Open in App

Solution

The correct option is B $3045 J$
Pictorial representation of conversion of ice at $- 10^{\circ} C$ into steam at $100^{\circ} C$ is given below
( $c_{i} =$ specific heat of ice, $c_{w} =$ specific heat of water)

Let $Q$ be the total heat required to convert ice into steam.
From the figure, we can deduce that,
$Q = Q_{1} + Q_{2} + Q_{3} + Q_{4}$
$Q = m c_{i} Δ θ + m L_{f} + m c_{w} Δ θ + m L_{v}$
Substituting given values,
$\Rightarrow Q = 1 \times 0.5 \times [0 - (- 10)] + 1 \times 80 + 1 \times 1 \times (100 - 0) + 1 \times 540$
$∴ Q = 725 cal$
Hence, work done in converting ice (at $- 10^{\circ} C$ ) into steam (at $100^{\circ} C$ ) is given by,
$W = J Q = 4.2 \times 725 = 3045 J$
Thus, option (b) is the correct answer.

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}

Q. The amount of heat required to convert

5 g

of ice at

0^{\circ} C

5 g

of steam at

100^{\circ} C

is -
[Latent heat of vaporization and fusion are

L_{v} = 540 {cal g}^{- 1}

and

L_{f} = 80 {cal g}^{- 1}

]

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

]

600 g

of ice at

253 K

is mixed with

0.08 kg

of steam at

100^{\circ} C

. Latent heat of vaporization of steam

= 540 cal/g

. Latent heat of fusion of ice

= 80 cal/g

. Specific heat of ice

= 0.5

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

. The resultant 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

)

Join BYJU'S Learning Program

Explore more

Specific Heat Capacity

Standard XI Physics

Join BYJU'S Learning Program

Work done in converting one gram of ice at −10∘C into steam at 100∘C is - [Given, specific heat of ice ci=0.5 cal g−1 ∘C−1, specific heat of water cw=1 cal g−1 ∘C−1, latent heat of fusion of ice Lf=80 cal/g, & latent heat of vaporization of steam Lv=540 cal/g]