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\( 1 \) gram of ice is mixed with \( 1 \) gram of steam. At thermal equilibrium, the temperature of the
mixture is
Options:
mixture is
Solution:
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Verified Answer
The correct answer is:
\( 100^{\circ} \mathrm{C} \)
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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