Question

10gm of ice at 0 C is kept in a calorimeter of water equivalent 10gm. how much heat should be supplied to the apparatus to evaporate the water thus formed? ( neglect loss of heat )

Answer 1

this is ur required result

Answer 2

Hey friend, Harish here.

Here is your answer:

Given that:

10 gm of ice at 0°C is kept in calorimeter of water 10 gm.

To find,

The amount of heat that must be supplied.

Solution:

First we must calculate the energy needed for ice at 0°C to Convert into water at same temperature. 

So,  We know that,

$Q_{(ice - water)} = amount\ of\ ice \times L_{f}\ of \ water$

Here , L v - Latent Heat of Fusion of water = 80 cal/gm 

⇒  $Q_{(ice - water)} = 10 g \times 80\ cal/gm = 800 cal$

Now it has to convert from 0°C water to 100°C of water.

Then,

⇒ $Q_{(0-100)} = [Amount\ of \ Water + Equivalent\ of\ water](S) (T_{f} - T_{i})$

Here , S- Specific Heat of water = (1 cal /gm k) , And Change in temperature is (T f) - (T i)

Then,

⇒ $Q_{(0-100)} = (10+10)(1)(100) = 20 \times 100 = 2000\ cal.$

Now, From 100°C water to Vapor at same temperature. 

$Q_{(Water-Vapour)} = (Amount\ of\ water)\times L_{v}$

Here Lv = 540 cal/gm.

⇒ $Q_{(Water-Vapour)}= 10 \times 540 = 5400\ cal$

Then,

⇒ $Total\ Heat\ needed = Sum\ of\ all\ energy\ used.$

⇒ $800+2000+5400 = 8200\ cal$

$\bold{Therefore\ total\ heat\ required\ is\ 8200\ cal}$
_____________________________________________________

Hope my answer is helpful to you.