Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Please solve this question. This question is from calorimetry.

Please solve this question. This question is from calorimetry.

Question Image
Grade:12th pass

1 Answers

Arun
25763 Points
2 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}
 
 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free