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

A volume of 120 ml of drink (half alcohol + half water by mass) originally at a temperature of 25°c is cooled by adding 20 gm of ice at 0°c. If all the ice melts, find the final temperature of the drink. (Density of drink = 0.833 gm/cc, specific heat of alcohol = 0.6 cal/gm.°c, specific heat of water = 1 cal/gm.°c, specific latent heat of ice = 80 cal/gm)

A volume of 120 ml of drink (half alcohol + half water by mass) originally at a temperature of 25°c is cooled by adding 20 gm of ice at 0°c. If all the ice melts, find the final temperature of the drink. (Density of drink = 0.833 gm/cc, specific heat of alcohol = 0.6 cal/gm.°c, specific heat of water = 1 cal/gm.°c, specific latent heat of ice = 80 cal/gm)

Grade:11

1 Answers

Eshan
askIITians Faculty 2095 Points
3 years ago
Let the final temperature beT.
Heat gained by ice=Heat lost by alcohol+water system
\implies (20)(80)+(20)(1)(T-0)=(120\times 0.833)(\dfrac{1}{2}(0.6)(25-T)+\dfrac{1}{2}(1)(25-T))Solving the equation gives,T=4^{\circ}C

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