Question

One gram of ice ( 0° C ) is mixed with one gram of steam ( 100° C ). After thermal equilibrium, the temperature of the mixture is

• A. C
• B. C
• C. C
• D. C

Answer 1

The correct option is B

C

1 gram of ice at 0° C takes 80 calories to change to 1 gram of water at 0° C. Also, the specific heat of water is 1 calorie/gram °C, i.e, 1 gram of water takes up 1 calorie in order to raise its temperature by 1 degree. But, the latent heat of vaporization of water is 540 cal/g/°C, i.e, 1 gram of steam releases 540 cal to become 1 gram of water at 100° C. Thus,
Amount of heat absorbed to convert 1 gram of ice at 0° C to 1 gram of water at 100° C = 180 cal
is less than
Amount of heat released during conversion of 1 gram of steam at 100° C to 1 gram of water at 100° C = 540 cal
therefore, thermal equillibrium is reached at 100° C