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Calorimetry

10 g ice at

Question

10 g ice at %%0^{\circ} C $i s c o n v e r t e d i n t o s t e a m a t$ 100^{\circ} C $. F i n d t o t a l h e a t r e q u i r e d ($ L_f = 80 cal/G $,$ S_w = 1 cal/g ^{\circ} C $, L_{v} = 540 c a l / g$ )

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Solution

Heat will be required to first melt the ice, then convert water to boiling point and then conversion to steam.

Hence,

$E_{t o t a l} = 10 (80 + 540 + 1 (100)) = 7200 c a l$

Answer is $7200 c a l$

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Similar questions

Find the amount of heat in calories required to convert 10g of ice at –10°C to water at 10°C

[C of ice = 0.5 cal/g°C, C of water = 4.2J/g°C = 1 cal/g°C, L of ice = 80cal/g]

100^{\circ} C

- 10^{\circ} C

= 0.5 c a l g^{- 1}^{\circ} C^{- 1}

= 540 c a l g^{- 1},

= 80 c a l g^{- 1}

10 kg

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

80^{\circ} C

0^{\circ} C

80 cal/g

- 10^{\circ} C

100^{\circ} C

c_{i} = 0.5 {cal g}^{- 1}^{\circ} C^{- 1}

c_{w} = 1 {cal g}^{- 1}^{\circ} C^{- 1}

L_{f} = 80 cal/g

L_{v} = 540 cal/g

1 g

- 10^{\circ} C

100^{\circ} C

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