One gram of ice is mixed with one gram of steam. At thermal equilibrium the temperature of mixture is
  • a)
    0°C
  • b)
    50°C
  • c)
    80°C
  • d)
    100°C
Correct 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
8 Answers
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.

Komal Srivastava answered 3 weeks ago
The correct answer is option number `D'

Manjula Sandala answered 13 hours ago
100 degrees

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