Question

10 g of ice at -1${}^o$C is added to 10 g of water at 85${}^o$C. What is the final temperature and amount of ice left in the system ?(System is kept inside an ideal insulator)

• A. 2.25${}^{o}C$, 0 g
• B. 0${}^o$, 2 g
• C. 37.5${}^o$, 0 g
• D. 37.5${}^o$, 5 g

Answer 1

The correct option is A 2.25${}^{o}C$, 0 g

Heat required to melt the $10g$ of ice at $0^{o}C$,

$Q=mL=10\times 80=800cal$,

Now heat given by $10g$ of water in cooling down from ${85}^{o}C$ to $0^{o}C$,

$Q^{\prime }=m_w{c}_w\Delta t=10\times 1\times (85-0)=850cal$

As $Q^{\prime }>Q$, therefore total $10g$ of ice will melt i.e. ice left in the system $=0g$.

Heat required to raise the temperature of $10g$ of ice from $-1^{o}C$ to $0^{o}C$,

$Q^{\prime \prime }=m_i{c}_i\Delta t^{\prime }=10\times 0.5\times (0+1)=5cal$

Heat required to melt the $10g$ of ice at $0^{o}C$,

$Q=mL=10\times 80=800cal$,

The remaining heat $Q^{\prime \prime \prime }=850-(800+5)=45cal$, will increase the temperature of $10+10=20g$ of water. Let this final temperature is $t$,

Hence, $Q^{\prime \prime \prime }=m_w^{\prime }{c}_w(t-0)$

$45=20\times 1\times t$

$t=45/20={2.25}^{o}C$