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
- A \( 0^{\circ} \mathrm{C} \)
- B \( 100^{\circ} \mathrm{C} \)
- C \( 50{ }^{\circ} \mathrm{C} \)
- D \( 55^{\circ} \mathrm{C} \)
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 \(100^{\circ} \mathrm{C}\).
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 \(100^{\circ} \mathrm{C}\).
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