Question

$50g$ of ice at 0 C is mixed with $50g$ of water at 20 C.The resultant temperature of the mixture would be

• A. 10 C
• B. 0 C
• C. -10 C
• D. -35 C

Answer 1

The correct option is B 0 C

$50g$ of Ice at $0^{0}C$ has a latent heat of

$Q=m\times L=50\times 80$

$=4000cal$

Now for water to reach $0^{0}C$ without changing its state

Heat released by water = $m{C}_p\Delta T$

= $50\times 1\times 20$

= $1000cal$

As the heat to be removed from water is less than the latent heat of $50g$ of ice, the resultant mixture stays at $0^{0}C$ temperature.