# Please solve this question. This question is from calorimetry.

Arun
25750 Points
5 years ago
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}