Calculate the mass of steam at 100-C needed to just melt 320 g of ice at -10-C- The specific heat of ice -0-5 cal g-1-C-1- latent heat of vaporization of steam -540 cal g-1- specific heat of fusion of ice -80 cal g-1-

Heat required =320×12× 10+320× 80 =1600+25600 =27200cal Let the mas os steam required = m m× 540=27200 ⇒ m=50.370gm ...

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Standard XII

Physics

Zeroth Law of Thermodynamics

Calculate the...

Question

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 $= 80 c a l g^{- 1}$ .

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Solution

Heat required $= 320 \times \frac{1}{2} \times 10 + 320 \times 80 = 1600 + 25600 = 27200 c a l$
Let the mas os steam required = $m$
$m \times 540 = 27200 \Rightarrow m = 50.370 g m$

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Similar questions

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

]

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. Calculate the heat required to convert 

3 k g

of ice

a t - 12^{\circ} C

kept in a calorimeter to steam at

100^{\circ}

at atmospheric pressure. Given : specific heat of ice

= 2100 \times 10^{5} J {k g}^{- 1} K^{- 1}

, specific heat of water

= 4186 \times 10^{3} J {k g}^{- 1} K^{- 1}

, latent heat of fusion of

c e = 3.35 \times 10^{5} J {k g}^{- 1}

and latent heat of steam

= 2.256 \times 10^{6} J {k g}^{- 1}

Q. The amount of heat required to convert

1

g of ice at

- 10^{0} C

into steam at

100^{0} C

is (approximately): (specific heat of water

=

4.187

J/g

^{0}

C, latent heat of fusion

=

334

J/g , latent heat of vaporization

=

2260

J/g, specific heat of ice

=

2.03

J/g) :

Calculate the amount of heat required to convert 1.00 kg of ice at -10°C at normal pressure to steam at 100°C. Specific heat capacity of ice = 2100 J kg⁻¹ K⁻¹ , latent heat of fusion of ice = 3.36 × 10⁵ J kg⁻¹ , specific heat capacity of water = 4200 J kg ⁻¹ K⁻¹ and latent heat of vaporization of water = 2.25 × 10⁶ J kg⁻¹

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Calculate the mass of steam at 100∘C needed to just melt 320 g of ice at −10∘C. The specific heat of ice =0.5calg−1∘C−1, latent heat of vaporization of steam =540calg−1, specific heat of fusion of ice =80calg−1.

Heat required =320×12×10+320×80=1600+25600=27200calLet the mas os steam required = mm×540=27200⇒m=50.370gm

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 $= 80 c a l g^{- 1}$ .

Heat required $= 320 \times \frac{1}{2} \times 10 + 320 \times 80 = 1600 + 25600 = 27200 c a l$
Let the mas os steam required = $m$
$m \times 540 = 27200 \Rightarrow m = 50.370 g m$