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Hess' Law

How much ener...

Question

How much energy is needed to convert $100 g$ of ice at $263 K$ to liquid water at a temperature of $283 K$ ?
$C_{i c e} = 0.49 c a l / (g^{o} C)$
$C_{w a t e r} = 1.00 c a l / (g^{o} C)$
$Δ H_{f u s} = 79.8 c a l / g$
$Δ H_{v a p} = 540 c a l / g$

9470 c a l

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3288 c a l

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2288 c a l

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727.3 c a l

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Solution

The correct option is A $9470 c a l$
ice $\to$ liquid water
$C_{i c e} = 0.49 c a l / (g^{o} C)$
$C_{w a t e r} = 1.00 c a l / (g^{o} C)$
$Δ H_{f u s} = 79.8 c a l / g$
$Δ H_{v a p} = 540 c a l / g$
ice $\to$ melt (fusicm) $\to$ water
$(263 k)$ At $0^{\circ} c (237 k)$ $283 k$
ENERGY $= \int_{263}^{273} C_{i c e} d t + Δ H_{f u s} + \int_{273}^{283} C_{w a t e r} d t$
$= 0.49 (273 - 263) + 79.8 + 1 (283 - 273)$
$= 94.7 c a l g$
For 100 g ice energy
$= 94.7 \times 100 c a l$
$= 9470 c a l$

Suggest Corrections

Similar questions

100 g

Δ H_{v} = 540 c a l / g

Δ H_{f} = 80 c a l / g

C_{w a t e r} = 1 c a l / g^{o} C

150

50

32^{o}

5^{o}

= 0.4 J / g^{o} C

= 4.2 J / g^{o} C

= 330 J / g

100^{o} C

(L_{s t e a m} = 540 c a l / g, L_{i c e} = 80 c a l / g; S_{w a t e r} = 1 c a l / g^{o} C)

C_{i c e} = 0.53 c a l / g -^{\circ} C

C_{w a t e r} = 1.0 c a l / g -^{\circ} C

(L_{f})_{w a t e r} = 80 c a l / g

(L_{v})_{w a t e r} = 529 c a l / g

0^{\circ} C

100 g

25^{o} C

- 10^{o} C

50

i c e = 0.5 c a l / g^{o} C

= 80 c a l / g)

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