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# One gram of ice is mixed with one gram of steam. At thermal equilibrium the temperature of mixture isa)0°Cb)50°Cc)80°Cd)100°CCorrect answer is option 'D'. Can you explain this answer? Related Test: Test: Heat - 1

## Class 7 Question

By Jitender Singh · Jan 14, 2020 ·Class 7
Alka Ekka answered Jan 10, 2020
Heat required melting 1g of ice at 0 degree to water at 0 degree = 80 cal
Heat required to raise the temperature of 1 g of water from 0 degree to 100 degree = 1 x 1 x 100 =100cal
Total heat required for maximum temperature of 100 = 180 cal
1 g of steam gives 540 cal so the temperature of mixture will be 100 degrees.

Bipin Singh answered Jul 24, 2020

Komal Srivastava answered Jul 22, 2020
The correct answer is option number `D'

Prableen Kaur answered Jan 14, 2020

Subhi Prajapati answered Aug 05, 2020
100

Dushyant Beda answered 2 weeks ago
100#â€°100

Manjula Sandala answered Aug 06, 2020
100 degrees

Chacko Babym answered Aug 09, 2020

Preet Bahedia answered Jul 04, 2020
Option

Sujata Panda answered Nov 21, 2019
B

Urvashi Saini answered 2 weeks ago
1gram ice hai to or one gram of steam mixed of all than answer of 0.C