Question

540 g of ice at $0^{o}C$ is mixed with 540 g of water at ${80}^{o}C$. The final temperature of the mixture is

• A. 0°C
• B. 40°C
• C. 80°C
• D. Less than 0°C

Answer 1

The correct option is A

0°C

Heat taken by ice to melt at $0^{o}C$ is
$Q_1=mL=540\times 80=43200cal$
Heat given by water to cool upto $0^{o}C$ is
$Q_2=ms△\theta =540\times 1\times (80-0)=43200cal$
Hence heat given by water is just sufficient to melt the whole ice and final temperature of mixture is $0^{o}c$

Short trick: for these type of frequentlu asked questions you can remember the following formula

${\theta }_{mix}=\frac{m_w{\theta }_w-\frac{m_i{L}_i}{c_w}}{m_i+m_w}$(See theory for more details)
If $m_w=m_{i}then{\theta }_{mix}=\frac{{\theta }_w-\frac{L_i}{c_w}}{2}=\frac{80-\frac{80}{1}}{2}=0^{o}C$