KCET2015PhysicsMotion in Two Dimensions
( 1 ) gram of ice is mixed with ( 1 ) gram of steam. At thermal equilibrium, the temperature of the mixture is
Options
- A( 0^ C )
- B( 100^ C )
- C( 50 ^ C )
- D( 55^ C )
Correct answer
B. ( 100^ C )
Step-by-step solution
Given, 1 gram of ice is mixed with 1 gram of steam. Now, heat required to melt 1 gram of ice at 0^ C to water =80 cal Heat required to raise the temperature of 1 gram of water from 0^ C to 100^ C =100 cal So, total heat required for maximum temperature of 100^ C =180 cal But temperature cannot be more than 100^ C . Therefore, temperature of mixture is 100^ C .