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

1gram of ice is mixed with 1gram of steam. At themeal equllibrium ,the temperature of mixture is

1gram of ice is mixed with 1gram of steam. At themeal equllibrium ,the temperature of mixture is
 

Grade:12th pass

1 Answers

Arun
25763 Points
2 years ago

1 gm of ice requires 334 joules of heat to change it into water as we know specific heat of water is 4.2J/gk

1 gm of water is converted into steam by giving 2230J/g of heat energy specific heat of steam is 1.996kj/kg k.

For attaining equilibrium temperature= ice to water latent heat + water latent heat to change to 0degree to 100degree change in temperature /latent heat of vaporization

Mass of steam =334+ 420/2230=0.338g.

Q=m×c×t

80 degree Celsius final temperature

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