CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

One gram of ice at 0C is mixed with one gram of steam at 100C . At thermal equilibrium the temperature of mixture is


A
0°C
loader
B
100°C 
loader
C
55°C  
loader
D
80°C
loader

Solution

The correct option is A 100°C 

Heat required to melt 1 g of ice of 0C to water at 0C = 1 × 80 cal.

Heat required to raise temperature of 1 g of water from 0C to 100C = 1 × 1 × 100 = 100 cal

Total heat required for maximum temperature of 100C = 80 + 100 = 180 cal

As one gram of steam gives 540 cal of heat when it is converted to water at 100°C, therefore, temperature of the mixture would be 100C.


Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image