Question

$10$g of ice at $-{20}^o$C is dropped into a calorimeter containing $10$g of water at ${10}^o$C; the specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain.

• A. $20$g of water
• B. $20$g of ice
• C. $10$g ice and $10$g water
• D. $5$g ice and $15$g water

Answer 1

The correct option is C $10$g ice and $10$g water

Final temperature of mixture can be greater than $0^{o}C$ or less than $0^{o}C$.

1) Final temperature greater than $0^{o}C$

As all ice in caloriemeter absorbs heat and rises to ${10}^{o}C$, Ice at $0^{o}C$ absorbs latent heat and becomes water at $0^{o}C$

$△\theta =$ required heat $=(10\times 0.5\times 20)+(10\times 80)+10\times 1\times \theta$

Heat lost by water $=10(10-\theta )=100-10\theta$

Required heat $>$ heat lost

2) Final temperature less than $0^{o}C$

Let the final temperature be $-{\theta }^{o}C$

Heat released by water to attain $-{\theta }^{o}C$ $=10\times 1\times 10+10\times 80+10\times 0.5\times \theta$

Heat absorbed by ice $=10\times 0.5(10-\theta )$

Since heat absorbed by ice less than heat to be released by water for all temperatures between $0^{o}C$ and $-{10}^{o}C$. The system can't attain a temperature less than $0^{o}C$.

3) So, final temperature of mixture can neither greater than $0^{o}C$ nor less than $0^{o}C$. So, the caloriemeter will contain $10g$ of ice and $10g$ of water and the final temperature will be $0^{o}C$