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Byju's Answer

Standard IX

Chemistry

Cooling from Evaporation

The heat requ...

Question

The heat required (in $c a l$ ) to change 10 g ice at $0^{o} C$ to steam at $100^{o} C$ is:
[ Heat of fusion and heat of vaporization for $H_{2} O$ are $80 c a l / g$ and $540 c a l / g$ respectively. Specific heat of water is $1 c a l / g$ ]

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Solution

The heat required for the melting of ice is $10 \times 80 = 800 c a l$

The heat required for evaporation of water is $10 \times 540 = 5400 c a l$

The heat required for changing the temperature of water from $0^{0} C t o 100^{0} C$ is $10 \times 1 \times 100 = 1000 c a l$

Total heat required for the entire process is $800 + 5400 + 1000 = 7200 c a l$

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Similar questions

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. The amount of heat required to convert

5 g

of ice at

0^{o} C

into water at

40^{o} C

is (Take heat of fusion

= 80 c a l / g

, specific heat

= 1 c a l / g

) :

Q. The amount of heat required to raise the temperature of

75 g

of ice at

0^{o} C

to water at

10^{o} C

is:

[latent heat of fusion of ice is

80 c a l / g

, specific heat of water is

1 c a l / g^{o} C

]

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

]

500 g

of water and

100 g

of ice at

0^{o} C

are in a calorimeter whose water equivalent is

40 g

10 g

of steam at

100^{o} C

added to it. Then water in the calorimeter is (Latent heat of ice

= 80 c a l / g

, Latent heat of steam

= 540 c a l / g

)

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The heat required (in cal) to change 10 g ice at 0oC to steam at 100oC is:[ Heat of fusion and heat of vaporization for H2O are 80cal/g and 540cal/g respectively. Specific heat of water is 1cal/g]

The heat required for the melting of ice is 10×80=800calThe heat required for evaporation of water is 10×540=5400 calThe heat required for changing the temperature of water from 00C to 1000C is 10×1×100=1000calTotal heat required for the entire process is 800+5400+1000=7200cal

The heat required (in $c a l$ ) to change 10 g ice at $0^{o} C$ to steam at $100^{o} C$ is:
[ Heat of fusion and heat of vaporization for $H_{2} O$ are $80 c a l / g$ and $540 c a l / g$ respectively. Specific heat of water is $1 c a l / g$ ]