The amount of heat required to convert 1 gm of Ice at 0 - to 1 gm of steam at 100 -C-

The correct option is A 720 calSolution Heat required = Mass of ice [latent heat of #160;fusion + 8 profit of heat × temp change +Latent heat of b ...

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Standard X

Physics

Latent Heat

The amount of...

Question

The amount of heat required to convert $1$ gm of Ice at $0^{\circ}$ to $1$ gm of steam at $100^{\circ} C$ .

720

cal

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640

cal

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540

cal

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80

cal

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Solution

The correct option is A $720$ cal
Solution Heat required = Mass of ice [latent heat of
fusion + 8 profit of heat $\times$ temp change +
Latent heat of boiling ]
$\Rightarrow Δ H = 1 g [L_{f} + 8 (Δ T) + L_{v}]$
$Δ H = 1 g [80 c a l / g + 1 c a l / g (100 - 0) + 540 c a l g]$
$Δ H = [80 + 100 + 540] = 7200$
Option - A is correct

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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. 1 g of a steam at

100^{_{-}^{0}} C

melt how much ice at

0^{_{-}^{0}} C

? (Latent heat of ice = 80 cal/gm and latent heat of steam = 540 cal/gm)-

20 gm

ice at

- 10^{\circ} C

is mixed with

m gm

steam at

100^{\circ} C

. The minimum value of

m

so that finally all ice and steam converts into water at

0^{\circ} C

(

Use: specific heat and latent heat as

C_{i c e} = 0.5 cal/gm^{\circ} C, C_{water} = 1 cal/gm^{\circ} C, L_{f u s i o n}

= 80 cal/gm

and

L_{v a p o r}

= 540 cal/gm

)

Q. Heat required to convert one gram of ice at

0^{_{-}^{0}} C

into steam at

100^{_{-}^{0}} C

is (given

L_{s t e a m}

= 536 cal/gm)-

Q. Amount of heat required to convert 10 g of ice at

0^{\circ}

Cof water at

20^{\circ}

C is

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The amount of heat required to convert 1 gm of Ice at 0∘ to 1 gm of steam at 100∘C.

The correct option is A 720 calSolution Heat required = Mass of ice [latent heat of fusion + 8 profit of heat × temp change +Latent heat of boiling ]⇒ΔH=1g[Lf+8(ΔT)+Lv]ΔH=1g[80cal/g+1cal/g(100−0)+540calg]ΔH=[80+100+540]=7200Option - A is correct

The amount of heat required to convert $1$ gm of Ice at $0^{\circ}$ to $1$ gm of steam at $100^{\circ} C$ .