Physics, asked by hansika09, 1 year ago

10 gram of ice at zero degree celsius is kept in a calorimeter of water equivalent 10 gram how much heat should be supplied to the apparatus apparatus to evaporate the water has formed

Answers

Answered by Anonymous
7

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}


Similar questions