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KCET · Physics · Motion In Two Dimensions

\( 1 \) gram of ice is mixed with \( 1 \) gram of steam. At thermal equilibrium, the temperature of the
mixture is

  1. A \( 0^{\circ} \mathrm{C} \)
  2. B \( 100^{\circ} \mathrm{C} \)
  3. C \( 50{ }^{\circ} \mathrm{C} \)
  4. D \( 55^{\circ} \mathrm{C} \)
Verified Solution

Answer & Solution

Correct Answer

(B) \( 100^{\circ} \mathrm{C} \)

Step-by-step Solution

Detailed explanation

Given, 1 gram of ice is mixed with 1 gram of steam.
Now, heat required to melt 1 gram of ice at \(0^{\circ} \mathrm{C}\) to water \(=80\) cal Heat required to raise the temperature of \(1 \mathrm{gram}\) of
water from \(0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}=100\) cal
So, total heat required for maximum temperature of \(100^{\circ} \mathrm{C}=180\) cal
But temperature cannot be more than \(100^{\circ} \mathrm{C}\). Therefore, temperature of mixture is \(100^{\circ} \mathrm{C}\).